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5.1.1: A.1- Graphing Quadratic Equations Using Properties and Applications

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    94089
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    Learning Objectives

    By the end of this section, you will be able to:

    • Recognize the graph of a quadratic equation in 2-variables
    • Find the axis of symmetry and vertex of a parabola
    • Find the intercepts of a parabola
    • Graph quadratic equations using properties
    • Solve maximum and minimum applications
    Be Prepared

    Before you get started, take this readiness quiz.

    1. Graph the equation \(y=x^{2}\) by plotting points.
    2. Solve: \(2 x^{2}+3 x-2=0\).
    3. Evaluate \(-\frac{b}{2 a}\) when \(a=3\) and \(b=-6\).

    Recognizing the Graph of a Quadratic Equation (in 2 variables)

    Previously we very briefly looked at the equation \(y=x^{2}\). It was one of the first non-linear equations we looked at. Now we will graph equations of the form \(y=a x^{2}+b x+c\) if \(a \neq 0\). We call the kind of expression on the right hand side a quadratic expression.

    I THINK THE BELOW ISN'T CORRECT.. ISN'T A QUADRATIC EQUATION IN TWO VARIABLES POTENTIALLY QUADRATIC IN X AND Y--I propose deleting this?

    Definition \(\PageIndex{1}\)

    Let \(A, B\), and \(C\) be real numbers with \(A≠0\). We call

    \(A x^{2}+B x+C\),

    a quadratic expression (in \(x\)). We will call an equation \(y=A x^{2}+B x+C\) a quadratic equation in (in two variables) , or more specifically, an equation quadratic in \(x\).

    Recall that the graph of an equation is a picture of the solutions we get by using an identification of solutions of the equation with points on a coordinate plane. This is ideal in numerous ways, for example, our drawing instruments are not precise or our space is limited (which it always is!). In what follows (as we did with lines before) we will be happy to capture certain features of the graph.

    We graph can the quadratic equation \(y=x^{2}\) by plotting points. That is, we can recognize that \((-3,9)\) is a solution to the equation since \(9=(-3)^2\). Similarly, \((-2,4)\) is a solution because \(4=(-2)^2\). Continuing down the chart below, we can substitute the value \(a\) for \(x\) to find a value \(b\) for \(y\) so that when we substitute \((a,b)\) in for \((x,y)\) we have a solution. When we do this, we see that for each value we substitute in for \(x\) there is only one value for \(y\) that will do the trick. We proceed to find these solutions on the Cartesian plane:

    This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis of the plane runs from negative 2 to 6. The parabola has a vertex at (0, 0) and also passes through the points (-2, 4), (-1, 1), (1, 1), and (2, 4). To the right of the graph is a table of values with 3 columns. The first row is a header row and labels each column, x, y equals x squared, and the order pair x, y. In row 2, x equals negative 3, y equals x squared is 9 and the ordered pair x, y is the ordered pair negative 3, 9. In row 3, x equals negative 2, y equals x squared is 4 and the ordered pair x, y is the ordered pair negative 2, 4. In row 4, x equals negative 1, y equals x squared is 1 and the ordered pair x, y is the ordered pair negative 1, 1. In row 5, x equals 0, y equals x squared is 0 and the ordered pair x, y is the ordered pair 0, 0. In row 6, x equals 1, y equals x squared is 1 and the ordered pair x, y is the ordered pair 1, 1. In row 7, x equals 2, y equals x squared is 4 and the ordered pair x, y is the ordered pair 2, 4. In row 8, x equals 3, y equals x squared is 9 and the ordered pair x, y is the ordered pair 3, 9.

    We call this figure a parabola. Parabolas are figures that can be squeezed vertically and horizontally, flipped or rotated, and translated so as to lay directly on the one above. We will look at a few special types. Let’s practice graphing a parabola by plotting a few points.

    Example \(\PageIndex{2}\)

    Graph \(y=x^{2}-1\).

    Solution

    We will graph the equation by plotting points.

    Choose integer values for \(x\),
    substitute them into the equation
    and simplify to find \(y\).
    Record the values of the ordered pairs in the chart.

    CNX_IntAlg_Figure_09_06_002a_img.jpg
    Plot the points, and then connect
    them with a smooth curve. The
    result will be the graph of the
    equation \(y=x^{2}-1\).
    .
    Try It \(\PageIndex{3}\)

    Graph \(y=-x^{2}\).

    Answer
    This figure shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (0, 0).
    Try It \(\PageIndex{4}\)

    Graph \(y=x^{2}+1\).

    Answer
    This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (0, −1).

    All graphs of quadratic equations of the form \(y=a x^{2}+b x+c\) are parabolas that open upward or downward.

    CNX_IntAlg_Figure_09_06_003_img_new2.jpg CNX_IntAlg_Figure_09_06_003_img_new3.jpg

    \(y=ax^2+bx+c\)

    \(y=x^2+4x+3\)

    \(a>0\)

    opens upward

    \(y=ax^2+bx+c\)

    \(y=-x^2+4x+3\)

    \(a<0\)

    opens downward

    Notice that the only difference in the two functions is the negative sign before the quadratic term (\(x^{2}\) in the equation of the graph in Figure 9.6.6). When the quadratic term, is positive, the parabola opens upward, and when the quadratic term is negative, the parabola opens downward.

    Definition \(\PageIndex{5}\)

    Parabola Orientation

    For the graph of the quadratic equation \(y=a x^{2}+b x+c\), if

    This images shows a bulleted list. The first bullet notes that, if a is greater than 0, then the parabola opens upward and shows an image of an upward-opening parabola. The second bullet notes that, if a is less than 0, then the parabola opens downward and shows an image of a downward-opening parabola.

    Example \(\PageIndex{6}\)

    Determine whether each parabola described by the following opens upward or downward:

    a. \(y=-3 x^{2}+2 x-4\)

    b. \(y=6 x^{2}+7 x-9\)

    Solution

    a. Find the value of \(a\).

    Comparing \(y=a x^{2}+b x+c\) to \(y=-3 x^{2}+2 x-4\), we see that \(a=-3\), which is negative. The associated parabola opens downward.

    b. Find the value of \(a\).

    Comparing \(y=a x^{2}+b x+c\) to \(y=6 x^{2}+7 x-9\), we see that \(a=6\), which is positive. The associated parabola opens upward.

    Try It \(\PageIndex{7}\)

    Determine whether each parabola described by the following opens upward or downward:

    a. \(y=2 x^{2}+5 x-2\)

    b. \(y=-3 x^{2}-4 x+7\)

    Answer

    a. up

    b. down

    Try It \(\PageIndex{8}\)

    Determine whether each parabola described by the following opens upward or downward:

    a. \(y=-2 x^{2}-2 x-3\)

    b. \(y=5 x^{2}-2 x-1\)

    Answer

    a. down

    b. up

    Find the Axis of Symmetry and Vertex of a Parabola

    Look again at Figure 9.6.10. Do you see that we could fold each parabola in half and then one side would lie on top of the other? The ‘fold line’ is a line of symmetry. We call it the axis of symmetry of the parabola.

    We show the same two graphs again with the axis of symmetry.

    This image shows 2 graphs side-by-side. The graph on the left shows an upward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1) and passes through the points (negative 4, 3) and (0, 3). The equation of this parabola is x squared plus 4 x plus 3. The vertical line passes through the point (negative 2, 0) and has the equation x equals negative 2. The graph on the right shows an downward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7) and passes through the points (0, 3) and (4, 3). The equation of this parabola is negative x squared plus 4 x plus 3. The vertical line passes through the point (2, 0) and has the equation x equals 2. This image shows 2 graphs side-by-side. The graph on the left shows an upward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1) and passes through the points (negative 4, 3) and (0, 3). The equation of this parabola is x squared plus 4 x plus 3. The vertical line passes through the point (negative 2, 0) and has the equation x equals negative 2. The graph on the right shows an downward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7) and passes through the points (0, 3) and (4, 3). The equation of this parabola is negative x squared plus 4 x plus 3. The vertical line passes through the point (2, 0) and has the equation x equals 2.
    \(y=x^2+4x+3\) \(y=-x^2+4x+3\)

    The equation of the axis of symmetry can be derived by using the Quadratic Formula. We will omit the derivation here and proceed directly to using the result. The equation of the axis of symmetry of the graph of \(y=a x^{2}+b x+c\) is \(x=-\frac{b}{2 a}\).

    So to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula \(x=-\frac{b}{2 a}\).

    \(y=ax^2+bx+c\) \(y=ax^2+bx+c\)
    \(y=x^2+4x+3\) \(y=-x^2+4x+3\)
    \(x=-\dfrac{b}{2a}\) \(x=-\dfrac{b}{2a}\)
    \(x=-\dfrac{4}{2\cdot1}\) \(x=-\dfrac{4}{2(-1)}\)
    \(x=-2\) \(x=2\)

    Notice that these are the equations of the dashed blue lines on the graphs.

    The point on the parabola that is the lowest (parabola opens up), or the highest (parabola opens down), lies on the axis of symmetry. This point is called the vertex of the parabola.

    We can easily find the coordinates of the vertex, because we know it is on the axis of symmetry. This means its
    \(x\)-coordinate is \(-\frac{b}{2 a}\). To find the \(y\)-coordinate of the vertex we substitute the value of the \(x\)-coordinate into the quadratic equation.

    \(y=x^2+4x+3\) \(y=-x^2+4x+3\)
    axis of symmetry is \(x=-2\) axis of symmetry is \(x=2\)
    vertex is \((-2, __)\) vertex is \((2, __)\)
    \(y=x^2+4x+3\) \(y=-x^2+4x+3\)
    \(y=(-2)^2+4(-2)+3\) \(y=-(2)^2+4(2)+3\)
    \(y=-1\) \(y=7\)
    vertex is \((-2, -1)\) vertex is \((2, 7)\)

    Axis of Symmetry and Vertex of a Parabola

    The graph of the equation \(y=a x^{2}+b x+c\) is a parabola where:

    • the axis of symmetry is the vertical line \(x=-\frac{b}{2 a}\).
    • the vertex is a point on the axis of symmetry, so its \(x\)-coordinate is \(-\frac{b}{2 a}\)
    • the \(y\)-coordinate of the vertex is found by substituting \(x=-\dfrac{b}{2 a}\) into the quadratic equation.
    Example \(\PageIndex{9}\)

    For the graph of \(y=3 x^{2}-6 x+2\) find:

    a. the axis of symmetry

    b. the vertex

    Solution

    a.

      \(a=3, b=-6, c=2\)
    The axis of symmetry is the vertical line \(x=-\frac{b}{2 a}\).  
    Substitute the values \(a,b\) into the equation. \(x=-\dfrac{-6}{2 \cdot 3}\)
    Simplify. \(x=1\)
      The axis of symmetry is the line \(x=1\).

    b.

       
    The vertex is a point on the line of symmetry, so its \(x\)-coordinate will be \(x=1\). Find the \(y\)-coordinate so that \((1,y\) is a solution. \(y=3(1)^2-6(1)+2\)
    Simplify. \(y=3\cdot 1-6+2=-1\)
    The result is the \(y\)-coordinate. \(y=-1\)
      \((1,-1)\) is the solution of the equation on the axis of symmetry. The vertex is \((1,-1)\).
    Try It \(\PageIndex{10}\)

    For the graph of \(y=2 x^{2}-8 x+1\) find:

    a. the axis of symmetry

    b. the vertex

    Answer

    a. \(x=2\)

    b. \((2,-7)\)

    Try It \(\PageIndex{11}\)

    For the graph of \(y=2 x^{2}-4 x-3\) find:

    a. the axis of symmetry

    b. the vertex

    Answer

    a. \(x=1\)

    b. \((1,-5)\)

    Find the Intercepts of a Parabola

    When we graphed linear equations, we often used the \(x\)- and \(y\)-intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.

    Remember, at the \(y\)-intercept the value of \(x\) is zero. So to find the \(y\)-intercept, we substitute \(x=0\) into the equation.

    Let’s find the \(y\)-intercepts of the two parabolas shown in Figure 9.6.20.

    This image shows 2 graphs side-by-side. The graph on the left shows an upward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1) and passes through the points (negative 4, 3) and (0, 3). The vertical line is an axis of symmetry for the parabola, and passes through the point (negative 2, 0). It has the equation x equals negative 2. The equation of this parabola is x squared plus 4 x plus 3. When x equals 0, f of 0 equals 0 squared plus 4 times 0 plus 3. F of 0 equals 3. The y-intercept of the graph is the point (0, 3). The graph on the right shows an downward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7) and passes through the points (0, 3) and (4, 3). The vertical line is an axis of symmetry for the parabola and passes through the point (2, 0). It has the equation x equals 2. The equation of this parabola is negative x squared plus 4 x plus 3. When x equals 0, f of 0 equals negative 0 squared plus 4 times 0 plus 3. F of 0 equals 3. The y-intercept of the graph is the point (0, 3). This image shows 2 graphs side-by-side. The graph on the left shows an upward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1) and passes through the points (negative 4, 3) and (0, 3). The vertical line is an axis of symmetry for the parabola, and passes through the point (negative 2, 0). It has the equation x equals negative 2. The equation of this parabola is x squared plus 4 x plus 3. When x equals 0, f of 0 equals 0 squared plus 4 times 0 plus 3. F of 0 equals 3. The y-intercept of the graph is the point (0, 3). The graph on the right shows an downward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7) and passes through the points (0, 3) and (4, 3). The vertical line is an axis of symmetry for the parabola and passes through the point (2, 0). It has the equation x equals 2. The equation of this parabola is negative x squared plus 4 x plus 3. When x equals 0, f of 0 equals negative 0 squared plus 4 times 0 plus 3. F of 0 equals 3. The y-intercept of the graph is the point (0, 3).

    \(y=x^2+4x+3\)

    Look for a solution of the form \((0,b)\).

    \(b=0^2+4\cdot 0+3\)

    \(b=3\)

    The \(y\)-intercept is \((0,3)\).

    \(y=x^2+4x+3\)

    Look for a solution of the form \((0,b)\).

    \(b=-0^2+4\cdot 0+3\)

    \(b=3\)

    The \(y\)-intercept is \((0,3)\).

       

    An \(x\)-intercept results when the value of \(f(x)\) is zero. To find an \(x\)-intercept, we let \(f(x)=0\). In other words, we will need to solve the equation \(0=a x^{2}+b x+c\) for \(x\).

    \(\begin{aligned} y &=a x^{2}+b x+c \\ 0 &=a x^{2}+b x+c \end{aligned}\)

    Solving quadratic equations like this (one variable) is exactly what we have done earlier in this chapter!

    We can now find the \(x\)-intercepts of the two parabolas we looked at. First we will find the \(x\)-intercepts of the parabola whose equation is \(y=x^{2}+4 x+3\).

    \(y=x^{2}+4 x+3\)
    Let \(y=0\). \(\color{red}0\color{black}=x^{2}+4 x+3\)
    Factor. \(0=(x+1)(x+3)\)
    Use the Zero Product Property. \(x+1=0 \quad x+3=0\)
    Solve.  
      The \(x\)-intercepts are \((-1,0)\) and \((-3,0)\).

    Now we will find the \(x\)-intercepts of the parabola whose equation is \(y=-x^{2}+4 x+3\).

    \(y=-x^{2}+4 x+3\)
    Let \(y=0\). \(\color{red}0 \color{black}=-x^{2}+4 x+3\)
    This quadratic does not factor, so we use the Quadratic Formula. \(a=-1, b=4, c=3\)
       
       
    Simplify.  
       
      \(x=\frac{-2(2 \pm \sqrt{7})}{-2}\)
      \(x=2 \pm \sqrt{7}\)
      The \(x\)-intercepts are \((2+\sqrt{7}, 0)\) and \((2-\sqrt{7}, 0)\).

    We will use the decimal approximations of the \(x\)-intercepts, so that we can locate these points on the graph,

    \((2+\sqrt{7}, 0) \approx(4.6,0) \quad(2-\sqrt{7}, 0) \approx(-0.6,0)\)

    Do these results agree with our graphs? See Figure 9.6.34

    This image shows 2 graphs side-by-side. The graph on the left shows the upward-opening parabola defined by the function f of x equals x squared plus 4 x plus 3 and a dashed vertical This image shows 2 graphs side-by-side. The graph on the left shows the upward-opening parabola defined by the function f of x equals x squared plus 4 x plus 3 and a dashed vertical

    \(y=x^2+4x+3\)

    \(y\)-intercept is \((0,3)\)

    \(x\)-intercepts are \((-1,0)\) and \((-3,0)\)

    \(y=-x^2+4x+3\)

    \(y\)-intercept is \((0,3)\)

    \(x\)-intercepts are \((2+\sqrt7,0)\approx(4.6,0)\) and \((2-\sqrt7,0)\approx(-0.6,0)\)

       

    Find the Intercepts of a Parabola

    To find the intercepts of a parabola whose equation is \(y=a x^{2}+b x+c\):

    \(y\)-intercept

    Let \(x=0\) and solve for \(y\).

    \(x\)-intercepts

    Let \(y=0\) and solve for \(x\)

    Example \(\PageIndex{12}\)

    Find the intercepts of the parabola whose equation is \(y=x^{2}-2 x-8\).

    Solution
    To find the \(y\)-intercept, let \(x=0\) and solve for \(y\).  
       
      \(y=-8\)
      When \(x=0\), then \(y=-8\). The \(y\)-intercept is the point \((0,-8)\).
    To find the \(x\)-intercept, let \(y=0\) and solve for \(x\).  
       
    Solve by factoring. \(0=(x-4)(x+2)\)
       
      \(4=x \quad-2=x\)
      When \(y=0\), then \(x=4\) or \(x=-2\). The \(x\)-intercepts are the points \((4,0)\) and \((-2,0)\).
    Try It \(\PageIndex{13}\)

    Find the intercepts of the parabola whose equation is \(y=x^{2}+2 x-8\).

    Answer

    \(y\) -intercept: \((0,-8)\)

    \(x\) -intercepts: \((-4,0),(2,0)\)

    Try It \(\PageIndex{14}\)

    Find the intercepts of the parabola whose equation is \(y=x^{2}-4 x-12\).

    Answer

    \(y\) -intercept: \((0,-12)\)

    \(x\) -intercepts: \((-2,0),(6,0)\)

    In this chapter, we have been solving quadratic equations of the form \(a x^{2}+b x+c=0\). We solved for \(x\) and the results were the solutions to the equation.

    We are now looking at quadratic equations of the form \(y=a x^{2}+b x+c\). The graphs of these equations are parabolas. The \(x\)-intercepts of the parabolas occur where \(y=0\).

    The solutions of the quadratic equation resulting from replacing \(y\) with \(0\) are the \(x\) values of the \(x\)-intercepts.

    Earlier, we saw that quadratic equations have \(2, 1\), or \(0\) solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the functions give the \(x\)-intercepts of the graphs, the number of \(x\)-intercepts is the same as the number of solutions.

    Previously, we used the discriminant to determine the number of solutions of a quadratic function of the form \(a x^{2}+b x+c=0\). Now we can use the discriminant to tell us how many \(x\)-intercepts there are on the graph.

    This image shows three graphs side-by-side. The graph on the left shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola lies below the x-axis and the parabola crosses the x-axis at two different points. If b squared minus 4 a c is greater than 0, then the quadratic equation a x squared plus b x plus c equals 0 has two solutions, and the graph of the parabola has 2 x-intercepts. The graph in the middle shows a downward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola lies on the x-axis, the only point of intersection between the parabola and the x-axis. If b squared minus 4 a c equals 0, then the quadratic equation a x squared plus b x plus c equals 0 has one solution, and the graph of the parabola has 1 x-intercept. The graph on the right shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola lies above the x-axis and the parabola does not cross the x-axis. If b squared minus 4 a c is less than 0, then the quadratic equation a x squared plus b x plus c equals 0 has no solutions, and the graph of the parabola has no x-intercepts.

    Before you to find the values of the \(x\)-intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

    Example \(\PageIndex{15}\)

    Find the intercepts of the parabola for the equation \(y=5 x^{2}+x+4\).

    Solution
        \(y=5 x^{2}+x+4\).  
    To find the \(y\)-intercept, let \(x=0\) and solve for \(y\).   \(y=5 0^{2}+0+4\).  
        \(y=4\)  
        When \(x=0\), then \(y=4\). The \(y\)-intercept is the point \((0,4)\).  
    To find the \(x\)-intercept, let \(f(x)=0\) and solve for \(x\).   \(0=5 x^{2}+x+4\).  
           
    Find the value of the discriminant to predict the number of solutions which is also the number of \(x\)-intercepts.  

    \(a=5,b=1,c=4\)

    \(b^2-4ac\)

    \(=1^2-4\cdot 5\cdot 4=-79\)

     
           
       

    Since the value of the discriminant is negative, there is no real solution to the equation.

    There are no \(x\)-intercepts.

     
    Try It \(\PageIndex{16}\)

    Find the intercepts of the parabola whose equation is \(y=3 x^{2}+4 x+4\).

    Answer

    \(y\)-intercept: \((0,4)\)

    no \(x\)-intercept

    Try It \(\PageIndex{17}\)

    Find the intercepts of the parabola whose equation is \(y=x^{2}-4 x-5\).

    Answer

    \(y\)-intercept: \((0,-5)\)

    \(x\)-intercepts: \((-1,0),(5,0)\)

    Graphing Quadratic Equations Using Properties

    Now we have all the pieces we need in order to graph a quadratic equation. We just need to put them together. In the next example we will see how to do this.

    Example \(\PageIndex{18}\)

    Graph \(y=x^{2}-6x+8\) by using its properties.

    Solution
    Determine whether the parabola opens upward or downward.

    Look \(a\) in the equation \(y=x^{2}-6x+8\)

    Since \(a\) is positive, the parabola opens upward.

    \(y=x^{2}-6x+8\)

    \(\color{red}{a=1, b=-6, c=8}\)

    The parabola opens upward.

    Find the axis of symmetry.

    \(y=x^{2}-6x+8\)

    The axis of symmetry is the line \(x=-\frac{b}{2 a}\).

    Axis of Symmetry

    \(x=-\frac{b}{2 a}\)

    \(x=-\frac{(-6)}{2 \cdot 1}\)

    \(x=3\)

    The axis of symmetry is the line \(x=3\).

    Find the vertex. The vertex is on the axis of symmetry. Substitute \(x=3\) into the function.

    Vertex

    \(y=x^{2}-6x+8\)

    the y-coordinate of the vertex \(=(\color{red}{3}\color{black}{)}^{2}-6(\color{red}{3}\color{black}{)}+8\)

    \(=-1\)

    The vertex is \((3,-1)\).

    Find the \(y\)-intercept. Find the point symmetric to the \(y\)-intercept across the axis of symmetry.

    We find solutions of the form \((0,y)\).

    We use the axis of symmetry to find a point symmetric to the \(y\)-intercept. The \(y\)-intercept is \(3\) units left of the axis of symmetry, \(x=3\). A point \(3\) units to the right of the axis of symmetry has \(x=6\).

    \(y\)-intercept

    The \(y\)-coordinate of the \(y\)-intercept

    \(=(\color{red}{0}\color{black}{)}^{2}-6(\color{red}{0}\color{black}{)}+8\)

    \(=8\)

    The \(y\)-intercept is \((0,8)\).

    The point symmetric to the \(y\)-intercept:

    The point is \((6,8)\).

    Find the \(x\)-intercepts. Find additional points if needed.

    We look for solutions of the form \((x,0)\).

    We can solve this quadratic equation by factoring.

    \(x\)-intercepts

    We substitute \(0\) for \(y\) in the equation:

    \(\color{red}{0}\color{black}{=}x^{2}-6x+8\)

    \(\color{red}{0}\color{black}{=}(x-2)(x-4)\)

    \(x=2 or x=4\)

    The \(x\)-intercepts are \((2,0)\) and \((4,0)\).

    Graph the parabola. We graph the vertex, intercepts, and the point symmetric to the \(y\)-intercept. We connect these \(5\) points to sketch the parabola. Screenshot (1).png
    Try It \(\PageIndex{19}\)

    Graph \(y=x^{2}+2x-8\) by using its properties.

    Answer
    This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The axis of symmetry, x equals negative 1, is graphed as a dashed line. The parabola has a vertex at (negative 1, negative 9). The y-intercept of the parabola is the point (0, negative 8). The x-intercepts of the parabola are the points (negative 4, 0) and (4, 0).
    Try It \(\PageIndex{20}\)

    Graph \(y=x^{2}-8x+12\) by using its properties.

    Answer
    This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 15. The axis of symmetry, x equals 4, is graphed as a dashed line. The parabola has a vertex at (4, negative 4). The y-intercept of the parabola is the point (0, 12). The x-intercepts of the parabola are the points (2, 0) and (6, 0).

    We list the steps to take in order to graph a quadratic function here.

    To Graph a Parabola Using Properties
    1. Determine whether the parabola opens upward or downward.
    2. Find the equation of the axis of symmetry.
    3. Find the vertex.
    4. Find the \(y\)-intercept. Find the point symmetric to the \(y\)-intercept across the axis of symmetry.
    5. Find the \(x\)-intercepts. Find additional points if needed.
    6. Graph the parabola.

    We were able to find the \(x\)-intercepts in the last example by factoring. We find the \(x\)-intercepts in the next example by factoring, too.

    Example \(\PageIndex{21}\)

    Graph \(y=x^{2}+6 x-9\) by using its properties.

    Solution
    \(y=x^{2}+6 x-9\)
    Since \(a\) is \(-1\), the parabola opens downward.  
      .
    To find the equation of the axis of symmetry, use \(x=-\frac{b}{2 a}\). \(x=-\frac{b}{2 a}\)
      \(x=-\frac{6}{2(-1)}\)
      \(x=3\)
     

    The axis of symmetry is \(x=3\).

    The vertex is on the line \(x=3\).

      .
    Find a solution of the form \((3,y)\). \(y=-x^{2}+6 x-9\)
      \(y=-3^{2}+6 \cdot 3-9\)
      \(y=-9+18-9\)
      \(y=0\)
      The vertex is \((3,0)\).
      .
    The \(y\)-intercept occurs when \(x=0\). Find a solution of the form \((0,y)\). \(y=-x^{2}+6 x-9\)
    xSubstitute \(x=0\). \(y=-0^{2}+6 \cdot 0-9\)
    Simplify. \(y=-9\)
    The point \((0,-9)\) is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is \((6,-9)\). .
      Point symmetric to the \(y\)-intercept is \((6,-9)\)
    The \(x\)-intercept occurs when \(y=0\). So we look for a solution of the form \(x,0)\) \(y=-x^{2}+6 x-9\)
    Find \(y=0\). .
    Factor the GCF. .
    Factor the trinomial. .
    Solve for \(x\). \(x=3\)
    Connect the points to graph the parabola. .
    Try It \(\PageIndex{22}\)

    Graph \(y=3 x^{2}+12 x-12\) by using its properties.

    Answer
    This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 15 to 10. The parabola has a vertex at (2, 0). The y-intercept (0, negative 12) is plotted as well as the axis of symmetry, x equals 2.
    Try It \(\PageIndex{23}\)

    Graph \(y=4 x^{2}+24 x+36\) by using its properties.

    Answer
    This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 30 to 20. The y-axis of the plane runs from negative 10 to 40. The parabola has a vertex at (negative 3, 0). The y-intercept (0, 36) is plotted as well as the axis of symmetry, x equals negative 3.

    For the graph of \(y=-x^{2}+6 x-9\), the vertex and the \(x\)-intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation \(0=-x^{2}+6x-9\) is \(0\), so there is only one solution. That means there is only one \(x\)-intercept, and it is the vertex of the parabola.

    How many \(x\)-intercepts would you expect to see on the graph of \(f(x)=x^{2}+4 x+5\)?

    Example \(\PageIndex{24}\)

    Graph \(y=x^{2}+4 x+5\) by using its properties.

    Solution
    \(y=x^{2}+4 x+5\)
    Since \(a\) is \(-1\), the parabola opens downward.  
      .
    To find the equation of the axis of symmetry, use \(x=-\frac{b}{2 a}\). .
      .
      .
     

    The equation of the axis of symmetry is \(x=-2\).

      .
    The vertex is on the line \(x=-2\).  
    Find \(y\) when \(x=-2\). \(y=x^{2}+4 x+5\)
      \(y=(-2)^{2}+4 (-2)+5\)
      \(y=4-8+5\)
      \(y=1\)
     

    The vertex is \((-2,1)\).

      .
    The \(y\)-intercept occurs when \(x=0\). \(y=x^{2}+4 x+5\)
    Find the solution of the form \((0,y)\). \(y=0^{2}+4 (0)+5\)
    Simplify. \(y=5\)
      The \(y\)-intercept is \((0,5)\).
    The point \((-4,5)\) is two units to the left of the line of symmetry. The point to units to the right of the line of symmetry is \((0,5)\. .
      Point symmetric to the \(y\)-intercept is \((-4,5)\).
    The \(x\)-intercept occurs when \(y=0\). .
    Find the solution of the form \((x,0)\). .
    Test the discriminant. \(a=1,b=4,c=5\)
      .
      .
      .
      .
    Since the value of the discriminant is negative, there is no real solution and so no \(x\)-intercept.  
    Connect the points to graph the parabola. You may want to choose two more points for greater accuracy. For example, you could try to find a solution of the form \((-1,y)\) which would lead to a solution \((-1,2\). .
    Try It \(\PageIndex{25}\)

    Graph \(y=x^{2}-2 x+3\) by using its properties.

    Answer
    This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 2 to 4. The y-axis of the plane runs from negative 1 to 5. The parabola has a vertex at (1, 2). The y-intercept (0, 3) is plotted as is the line of symmetry, x equals 1.
    Try It \(\PageIndex{26}\)

    Graph \(y=-3x^{2}-6 x-4\) by using its properties.

    Answer
    This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 2. The y-axis of the plane runs from negative 5 to 1. The parabola has a vertex at (negative 1, negative 2). The y-intercept (0, negative 4) is plotted as is the line of symmetry, x equals negative 1.

    Finding the \(y\)-intercept of the graph of an equation that looks like \(y=ax^2+bx+c\) is easy, isn't it? Finding the \(x\)-intercepts of such a graph can be more challenging. Sometimes we need to use the Quadratic Formula!

    Example \(\PageIndex{27}\)

    Graph \(y=2 x^{2}-4 x-3\) by using its properties.

    Solution
    \(y=2 x^{2}-4 x-3\)

    Since \(a\) is \(2\), the parabola opens upward.

    .
    To find the equation of the axis of symmetry, use \(x=-\frac{b}{2 a}\). \(x=-\frac{b}{2 a}\)
      \(x=-\frac{-4}{2 \cdot 2}\)
      \(x=1\)
      The equation of the axis of symmetry is \(x=1\).
    The vertex is on the line \(x=1\).  
    Find a solution of the form \((1,y)\). \(y=2 1^{2}-4 \cdot 1-3\)
      \(y=2-4-3\)
      \(y=-5)
      The vertex is \((1,-5)\).
    The \(y\)-intercept occurs when \(x=0\).  
    Find a solution of the form \((0,y)\). \(y=2 0^{2}-4 \cdot 0-3\)
    Simplify. \(y=-3\)
      The \(y\)-intercept is \((0,-3)\).
    The point \((0,-3)\) is one unit to the left of the line of symmetry. Point symmetric to the \(y\)-intercept is \((2,-3)\)
    The point one unit to the right of the line of symmetry is \((2,3)\).  
    The \(x\)-intercept occurs when \(y=0\).  
    Find a solution of the form \((x,0)\). .
    Use the Quadratic Formula.  
    Substitute in the values of \(a,b\) and \(c\). \(x=\frac{-(-4) \pm \sqrt{(-4)^{2}-4(2)(3)}}{2(2)}\)
    Simplify.  
    Simplify inside the radical. \(x=\frac{4 \pm \sqrt{40}}{4}\)
    Simplify the radical. \(x=\frac{4 \pm 2 \sqrt{10}}{4}\)
    Factor the GCF. \(x=\frac{2(2 \pm \sqrt{10})}{4}\)
    Remove common factors. \(x=\frac{2 \pm \sqrt{10}}{2}\)
    Write as two equations. \(x=\frac{2+\sqrt{10}}{2}, \quad x=\frac{2-\sqrt{10}}{2}\)
    Approximate the values. \(x \approx 2.5, \quad x \approx-0.6\)
      The approximate values of the \(x\)-intercepts are \((2.5,0)\) and \((-0.6,0)\).
    Graph the parabola using the points found. .
    Try It \(\PageIndex{28}\)

    Graph \(y=5 x^{2}+10 x+3\) by using its properties.

    Answer
    This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis of the plane runs from negative 4 to 4. The axis of symmetry, x equals negative 1, is graphed as a dashed line. The parabola has a vertex at (negative 1, negative 2). The y-intercept of the parabola is the point (0, 3). The x-intercepts of the parabola are approximately (negative 1.6, 0) and (negative 0.4, 0).
    Try It \(\PageIndex{29}\)

    Graph \(y=-3 x^{2}-6 x+5\) by using its properties.

    Answer
    This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The axis of symmetry, x equals negative 1, is graphed as a dashed line. The parabola has a vertex at (negative 1, 8). The y-intercept of the parabola is the point (0, 5). The x-intercepts of the parabola are approximately (negative 2.6, 0) and (0.6, 0).

    Solve Maximum and Minimum Applications

    Knowing that the vertex of a parabola is the lowest or highest point of the parabola gives us an easy way to determine the minimum or maximum value of a quadratic expression. Given a quadratic expression, with variable \(x\), say, we consider the equation in two variables \y\) equals the given expression. The y-coordinate of the vertex is the minimum value of that expression if the graph of the equation is a parabola that opens upward. It is the maximum value of the expression if the graph of the equation is a parabola that opens downward.

    This figure shows 2 graphs side-by-side. The left graph shows a downward opening parabola plotted in the x y-plane. An arrow points to the vertex with the label maximum. The right graph shows an upward opening parabola plotted in the x y-plane. An arrow points to the vertex with the label minimum.

    Optional: Minimum or Maximum Values of a Quadratic Expression

    Minimum and Maximum Values

    The \(y\)-coordinate of the vertex of the graph of a quadratic equation is the

    • minimum value of the quadratic expression if the parabola opens upward.
    • maximum value of the quadratic expression if the parabola opens downward.
    Example \(\PageIndex{30}\)

    Find the minimum or maximum value of the quadratic expression \(x^{2}+2 x-8\).

    Solution
    Consider the equation \(y=x^{2}+2 x-8\)
    Since \(a\) is positive, the parabola opens upward. The quadratic equation has a minimum.  
    Find the equation of the axis of symmetry. \(x=-\frac{b}{2 a}\)
      \(x=-\frac{2}{2 \times 1}\)
      \(x=-1\)
      The equation of the axis of symmetry is \(x=-1\).
    The vertex is on the line \(x=-1\). \(y=x^{2}+2 x-8\)
    Find the \(y\)-coordinate of the vertex. \(y=(-1)^{2}+2 (-1)-8\)
      \(y=1-2-8\)
      \(y=-9\)
      The vertex is \((-1,-9)\).
    Since the parabola has a minimum, the \(y\)-coordinate of the vertex is the minimum \(y\)-value of the quadratic equation. The minimum value of the quadratic is \(-9\) and it occurs when \(x=-1\).  
      .

    Show the graph to verify the result.

    The minimum value of the expression is \(-9\) and it occurs when \(x=-1\).

    Try It \(\PageIndex{31}\)

    Find the maximum or minimum value of the quadratic expression \(x^{2}-8 x+12\).

    Answer

    The minimum value of the quadratic expression is \(−4\) and it occurs when \(x=4\).

    Try It \(\PageIndex{32}\)

    Find the maximum or minimum value of the quadratic expression \(-4 x^{2}+16 x-11\).

    Answer

    The maximum value of the quadratic expression is \(5\) and it occurs when \(x=2\).

    We have used the formula

    \(h=-16t^2+v_0 t\)

    to calculate the height in feet, \(h\), of an object shot upwards into the air with initial velocity, \(v_{0}\), after \(t\) seconds (assuming the initial shot occurs at height \(0\)). The solutions to this equation in the context of the application are precisely the pairs \((t,h)\) so that \(h\) is the height of the object at time \(t\).

    This formula is a quadratic equation in \(t\), so its graph is a parabola. By solving for the coordinates of the vertex \((t,h)\), we can find how long it will take the object to reach its maximum height. Then we can calculate the maximum height.

    Example \(\PageIndex{33}\)

    The quadratic expression \(-16 t^{2}+176 t+4\) models the height of a volleyball hit straight upwards with velocity \(176\) feet per second from a height of \(4\) feet.

    a. How many seconds will it take the volleyball to reach its maximum height?

    b. Find the maximum height of the volleyball.

    Solution

    Let us consider the equation \(h=-16 t^{2}+176 t+4\) (here we have used \(h\) for 'height' instead of \(y\)). So again, the solutions of the equation are exactly the ordered pairs \((t,h)\) so that \(h\) is the height at time \(t\).

    Since \(a\) is negative, the parabola opens downward. The quadratic function has a maximum.

    a. Find the equation of the axis of symmetry.

    \(\begin{array}{l}{t=-\frac{b}{2 a}} \\ {t=-\frac{176}{2(-16)}} \\ {t=5.5}\end{array}\)

    The equation of the axis of symmetry is \(t=5.5\).

    The vertex is on the line \(t=5.5\).

    The maximum occurs when \(t=5.5\) seconds.

    b. Find the height at time \(t=5.5\). This is the same as searching for a solution of the form \((h,5.5)\).

    Use a calculator to simplify.

    \(h=488\)

    The vertex is \((5.5,488)\).

    Since the parabola has a maximum, the \(h\)-coordinate of the vertex is the maximum value of the quadratic expression.

    The maximum value of the quadratic expression is then \(488\) feet and it occurs when \(t=5.5\) seconds.

    After \(5.5\) seconds, the volleyball will reach its maximum height of \(488\) feet.

    E

    Exercise \(\PageIndex{34}\)

    Solve, rounding answers to the nearest tenth.

    The quadratic function \(h=-16 t^{2}+128 t+32\) is used to find the height of a stone thrown upward from a height of \(32\) feet at a rate of \(128\) ft/sec. How long will it take for the stone to reach its maximum height? What is the maximum height?

    Answer

    It will take \(4\) seconds for the stone to reach its maximum height of \(288\) feet.

    Exercise \(\PageIndex{35}\)

    A path of a toy rocket thrown upward from the ground at a rate of \(208\) ft/sec is modeled by the quadratic function of \(h=-16 t^{2}+208 t\). When will the rocket reach its maximum height? What will be the maximum height?

    Answer

    It will take \(6.5\) seconds for the rocket to reach its maximum height of \(676\) feet.

    Key Concepts

    • Parabola Orientation
      • For the graph of the quadratic equation \(y=a x^{2}+b x+c\), if
        • \(a>0\), the parabola opens upward.
        • \(a<0\), the parabola opens downward.
    • Axis of Symmetry and Vertex of a Parabola The graph of the equation \(y=a x^{2}+b x+c\) is a parabola where:
      • the axis of symmetry is the vertical line \(x=-\frac{b}{2 a}\).
      • the vertex is a point on the axis of symmetry, so its \(x\)-coordinate is \(-\frac{b}{2 a}\).
      • the \(y\)-coordinate of the vertex is found by substituting \(x=-\frac{b}{2 a}\) into the quadratic equation.
    • Find the Intercepts of a Parabola
      • To find the intercepts of a parabola whose equation is \(y=a x^{2}+b x+c\):
        • \(y\)-intercept
          • Let \(x=0\) and solve for \(y\).
        • \(x\)-intercepts
          • Let \(y=0\) and solve for \(x\).
    • How to graph a quadratic equation using properties.
      1. Determine whether the parabola opens upward or downward.
      2. Find the equation of the axis of symmetry.
      3. Find the vertex.
      4. Find the \(y\)-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
      5. Find the \(x\)-intercepts. Find additional points if needed.
      6. Graph the parabola.
    • Minimum or Maximum Values of a Quadratic Expression
      • The \(y\)-coordinate of the vertex of the graph of a quadratic equation formed by setting the expression (in \(x\)) equal to \(y\) is the
      • minimum value of the quadratic expression if the parabola opens upward.
      • maximum value of the quadratic expression if the parabola opens downward.

    Glossary

    quadratic expression (in x)
    An expression of the form \(ax^{2}+bx+c\), where \(a, b\), and \(c\) are real numbers and \(a≠0\), is called a quadratic expression (in \(x\)).
    quadratic equation (in x)
    An equation which can be written in the form \(y=ax^{2}+bx+c\), where \(a, b\), and \(c\) are real numbers and \(a≠0\) is a quadratic equation (in \(x\))

    5.1.1: A.1- Graphing Quadratic Equations Using Properties and Applications is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.