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Mathematics LibreTexts

10.5: Graphing Quadratic Equations

  • Page ID
    19000
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    By the end of this section, you will be able to:
    • Recognize the graph of a quadratic equation in two variables
    • Find the axis of symmetry and vertex of a parabola
    • Find the intercepts of a parabola
    • Graph quadratic equations in two variables
    • Solve maximum and minimum applications

    Before you get started, take this readiness quiz.

    1. Graph the equation \(y=3x−5\) by plotting points.
      If you missed this problem, review [link].
    2. Evaluate \(2x^2+4x−1\)when \(x=−3\)
      If you missed this problem, review [link].
    3. Evaluate \(−\frac{b}{2a}\) when \(a=13\) and b=\(\frac{5}{6}\)
      If you missed this problem, review [link].

    Recognize the Graph of a Quadratic Equation in Two Variables

    We have graphed equations of the form \(Ax+By=C\). We called equations like this linear equations because their graphs are straight lines.

    Now, we will graph equations of the form \(y=ax^2+bx+c\). We call this kind of equation a quadratic equation in two variables.

    definition: QUADRATIC EQUATION IN TWO VARIABLES

    A quadratic equation in two variables, where a,b,and c are real numbers and \(a\neq 0\), is an equation of the form \[y=ax^2+bx+c \nonumber\]

    Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations.

    Let’s look first at graphing the quadratic equation \(y=x^2\). We will choose integer values of x between −2 and 2 and find their y values. See Table.

    \(y=x^2\)
    x y
    0 0
    1 1
    \(−1\) 1
    2 4
    \(−2\) 4

    Notice when we let \(x=1\) and \(x=−1\), we got the same value for y.

    \[\begin{array} {ll} {y=x^2} &{y=x^2} \\ {y=1^2} &{y=(−1)^2} \\ {y=1} &{y=1} \\ \nonumber \end{array}\]

    The same thing happened when we let \(x=2\) and \(x=−2\).

    Now, we will plot the points to show the graph of \(y=x^2\). See Figure.

    This figure shows an upward-opening u shaped curve 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 lowest point on the curve is at the point (0, 0). Other points on the curve are located at (-2, 4), (-1, 1), (1, 1) and (2, 4).

    The graph is not a line. This figure is called a parabola. Every quadratic equation has a graph that looks like this.

    In Example you will practice graphing a parabola by plotting a few points.

    Example \(\PageIndex{1}\)

    \(y=x^2-1\)

    Answer

    We will graph the equation by plotting points.


    Choose integers values for x, substitute them into the equation and solve for y.
     
    Record the values of the ordered pairs in the chart. .
    Plot the points, and then connect them with a smooth curve. The result will be the graph of the equation \(y=x^2−1\) .

    Example \(\PageIndex{2}\)

    Graph \(y=−x^2\).

    Answer

    This figure shows a downward-opening u shaped curve 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 highest point on the curve is at the point (0, 0). Other points on the curve are located at (-2, -4), (-1, -1), (1, -1) and (2, -4).

    Example \(\PageIndex{3}\)

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

    Answer

    This figure shows an upward-opening u shaped curve 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 lowest point on the curve is at the point (0, 1). Other points on the curve are located at (-2, 5), (-1, 2), (1, 2) and (2, 5).

    How do the equations \(y=x^2\)and \(y=x^2−1\) differ? What is the difference between their graphs? How are their graphs the same?

    All parabolas of the form \(y=ax^2+bx+c\) open upwards or downwards. See Figure.

    This figure shows two graphs side by side. The graph on the left side shows an upward-opening u shaped curve 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 lowest point on the curve is at the point (-2, -1). Other points on the curve are located at (-3, 0), and (-1, 0). Below the graph is the equation y equals a squared plus b x plus c. Below that is the equation of the graph, y equals x squared plus 4 x plus 3. Below that is the inequality a greater than 0 which means the parabola opens upwards. The graph on the right side shows a downward-opening u shaped curve 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 highest point on the curve is at the point (2, 7). Other points on the curve are located at (0, 3), and (4, 3). Below the graph is the equation y equals a squared plus b x plus c. Below that is the equation of the graph, y equals negative x squared plus 4 x plus 3. Below that is the inequality a less than 0 which means the parabola opens downwards.

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

    Definition: PARABOLA ORIENTATION

    For the quadratic equation \(y=ax^2+bx+c\), if:

    The image shows two statements. The first statement reads “a greater than 0, the parabola opens upwards”. This statement is followed by the image of an upward opening parabola. The second statement reads “a less than 0, the parabola opens downward”. This statement is followed by the image of a downward opening parabola.

    Example\(\PageIndex{4}\)

    Determine whether each parabola opens upward or downward:

    1. \(y=−3x^2+2x−4\)
    2. \( y=6x^2+7x−9\)
    Answer
     

    .

    Since the “a” is negative, the parabola will open downward.

     

    .

    Since the “a” is positive, the parabola will open upward.

    Example\(\PageIndex{5}\)

    Determine whether each parabola opens upward or downward:

    1. \(y=2x^2+5x−2\)
    2. \(y=−3x^2−4x+7\)
    Answer
    1. up
    2. down

    Example \(\PageIndex{6}\)

    Determine whether each parabola opens upward or downward:

    1. \(y=−2x^2−2x−3\)
    2. \(y=5x^2−2x−1\)
    Answer
    1. down
    2. up

    Find the Axis of Symmetry and Vertex of a Parabola

    Look again at Figure. Do you see that we could fold each parabola in half and that 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 in red. See Figure.

    This figure shows an two graphs side by side. The graph on the left side 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 lowest point on the curve is at the point (-2, -1). Other points on the curve are located at (-3, 0), and (-1, 0). Also on the graph is a dashed vertical line that goes through the center of the parabola at the point (-2, -1). Below the graph is the equation of the graph, y equals x squared plus 4 x plus 3. The graph on the right side 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 highest point on the curve is at the point (2, 7). Other points on the curve are located at (0, 3), and (4, 3). Also on the graph is a dashed vertical line that goes through the center of the parabola at the point (2, 7). Below the graph is the equation of the graph, y equals negative x squared plus 4 x plus 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 \(y=ax^2+bx+c\) is x=\(−\frac{b}{2a}\).

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

    The figure shows the steps to find the axis of symmetry for two parabolas. On the left side the standard form of a quadratic equation which is y equals a x squared plus b x plus c is written above the given equation y equals x squared plus 4 x plus 3. The axis of symmetry is the equation x equals negative b divided by the quantity two times a. Plugging in the values of a and b from the quadratic equation the formula becomes x equals negative 4 divided by the quantity 2 times 1, which simplifies to x equals negative 2. On the right side the standard form of a quadratic equation which is y equals a x squared plus b x plus c is written above the given equation y equals negative x squared plus 4 x plus 3. The axis of symmetry is the equation x equals negative b divided by the quantity two times a. Plugging in the values of a and b from the quadratic equation the formula becomes x equals negative 4 divided by the quantity 2 times -1, which simplifies to x equals 2.
    Figure. Are these the equations of the dashed red lines?

    The point on the parabola that is on the axis of symmetry is the lowest or highest point on the parabola, depending on whether the parabola opens upwards or downwards. 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 \(−\frac{b}{2a}\). To find the y-coordinate of the vertex, we substitute the value of the x-coordinate into the quadratic equation.

    The figure shows the steps to find the vertex for two parabolas. On the left side is the given equation y equals x squared plus 4 x plus 3. Below the equation is the statement “axis of symmetry is x equals -2”. Below that is the statement “vertex is” next to the statement is an ordered pair with x-value of -2, the same as the axis of symmetry, and the y-value is blank. Below that the original equation is rewritten. Below the equation is the equation with -2 plugged in for the x value which is y equals -2 squared plus 4 times -2 plus 3. This simplifies to y equals -1. Below this is the statement “vertex is (-2, -1)”. On the right side is the given equation y equals negative x squared plus 4 x plus 3. Below the equation is the statement “axis of symmetry is x equals 2”. Below that is the statement “vertex is” next to the statement is an ordered pair with x-value of 2, the same as the axis of symmetry, and the y-value is blank. Below that the original equation is rewritten. Below the equation is the equation with 2 plugged in for the x value which is y equals negative the quantity 2 squared, plus 4 times 2 plus 3. This simplifies to y equals 7. Below this is the statement “vertex is (2, 7)”.

    Definition: AXIS OF SYMMETRY AND VERTEX OF A PARABOLA

    For a parabola with equation \(y=ax^2+bx+c\):

    • The axis of symmetry of a parabola is the line x=\(−\frac{b}{2a}\).
    • The vertex is on the axis of symmetry, so its x-coordinate is \(−\frac{b}{2a}\).

    To find the y-coordinate of the vertex, we substitute x=\(−\frac{b}{2a}\) into the quadratic equation.

    Example\(\PageIndex{7}\)

    For the parabola \(y=3x^2−6x+2\) find:

    1. the axis of symmetry and
    2. the vertex.
    Answer
    1. .
    The axis of symmetry is the line x=\(−\frac{b}{2a}\) .
    Substitute the values of a, b into the equation. .
    Simplify x=1
      The axis of symmetry is the line x=1
    2. .
    The vertex is on the line of symmetry, so its x-coordinate will be x=1  
    Substitute x=1 into the equation and solve for y. .
    Simplify .
    This is the y-coordinate. y=−1
    The vertex is (1,−1).

    Example \(\PageIndex{8}\)

    For the parabola \(y=2x^2−8x+1\) find:

    1. the axis of symmetry and
    2. the vertex.
    Answer
    1. x=2
    2. (2,−7)

    Example \(\PageIndex{9}\)

    For the parabola \(y=2x^2−4x−3\) find:

    1. the axis of symmetry and
    2. the vertex.
    Answer
    1. x=1
    2. (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 the figure below.

    This figure shows an two graphs side by side. The graph on the left side 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 vertex is at the point (-2, -1). Other points on the curve are located at (-3, 0), and (-1, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -2. Below the graph is the equation of the graph, y equals x squared plus 4 x plus 3. Below that is the statement “x equals 0”. Next to that is the equation of the graph with 0 plugged in for x which gives y equals 0 squared plus4 times 0 plus 3. This simplifies to y equals 3. Below the equation is the statement “y-intercept (0, 3)”. The graph on the right side 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 vertex is at the point (2, 7). Other points on the curve are located at (0, 3), and (4, 3). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2. Below the graph is the equation of the graph, y equals negative x squared plus 4 x plus 3. Below that is the statement “x equals 0”. Next to that is the equation of the graph with 0 plugged in for x which gives y equals negative quantity 0 squared plus 4 times 0 plus 3. This simplifies to y equals 3. Below the equation is the statement “y-intercept (0, 3)”.

    At an x-intercept, the value of y is zero. To find an x-intercept, we substitute \(y=0\) into the equation. In other words, we will need to solve the equation \(0=ax^2+bx+c\) for x.

    \[\begin{array} {ll} {y=ax^2+bx+c} \\ {0=ax^2+bx+c} \\ \nonumber \end{array}\]

    But solving quadratic equations like this is exactly what we have done earlier in this chapter.

    We can now find the x-intercepts of the two parabolas shown in Figure.

    First, we will find the x-intercepts of a parabola with equation \(y=x^2+4x+3\).

      .
    Let y=0 .
    Factor. .
    Use the zero product property. .
    Solve. .
      The x intercepts are (−1,0) and (−3,0).

    Now, we will find the x-intercepts of the parabola with equation \(y=−x^2+4x+3\).

      .
    Let y=0 .
    This quadratic does not factor, so we use the Quadratic Formula. .
    a=−1, b=4, c=3. .
    Simplify. .
    .
    ..
      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.

    \[\begin{array} {l} {(2+\sqrt{7},0) \approx (4.6,0)} & {(2−\sqrt{7},0) \approx (-0.6,0)}\\ \nonumber \end{array}\]

    Do these results agree with our graphs? See Figure.

    This figure shows an two graphs side by side. The graph on the left side 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 vertex is at the point (-2, -1). Three points are plotted on the curve at (-3, 0), (-1, 0), and (0, 3). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -2. Below the graph is the equation of the graph, y equals x squared plus 4 x plus 3. Below that is the statement “y-intercept (0, 3)”. Below that is the statement “x-intercepts (-1, 0) and (-3, 0)”. The graph on the right side 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 vertex is at the point (2, 7). Three points are plotted on the curve at (-0.6, 0), (4.6, 0), and (0, 3). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2. Below the graph is the equation of the graph, y equals negative x squared plus 4 x plus 3. Below that is the statement “y-intercept (0, 3)”. Below that is the statement “x-intercepts (2 plus square root of 7, 0) is approximately equal to (4.6, 0) and (2 minus square root of 7, 0) is approximately equal to (-0.6, 0).”

    Definition: FIND THE INTERCEPTS OF A PARABOLA

    To find the intercepts of a parabola with equation \(y=ax^2+bx+c\):

    \[\begin{array}{ll} {\textbf{y-intercept}}& {\textbf{x-intercept}}\\ {\text{Let} x=0 \text{and solve the y}}& {\text{Let} y=0 \text{and solve the x}}\\ \nonumber \end{array}\]

    Example \(\PageIndex{10}\)

    Find the intercepts of the parabola \(y=x^2−2x−8\).

    Answer
      .
    To find the y-intercept, let x=0 and solve for y. .
      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. .
     

    .

     
    When y=0, then x=4 or x=−2. The x-intercepts are the points (4,0) and (−2,0).
     

    Example \(\PageIndex{11}\)

    Find the intercepts of the parabola \(y=x^2+2x−8\).

    Answer

    y:(0,−8); x:(−4,0), (2,0)

    Example \(\PageIndex{12}\)

    Find the intercepts of the parabola \(y=x^2−4x−12\).

    Answer

    y:(0,−12); x:(6,0), (−2,0)

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

    We are now looking at quadratic equations in two variables of the form \(y=ax^2+bx+c\). The graphs of these equations are parabolas. The x-intercepts of the parabolas occur where y=0.

    For example:

    \[\begin{array}{cc} {\textbf{Quadratic equation}}&{\textbf{Quadratic equation in two variable}}\\ {}&{y=x^2−2x−15}\\ {x^2−2x−15}&{\text{Let} y=0, 0=x^2−2x−15}\\ {(x−5)(x+3)=0}&{0=(x−5)(x+3)}\\ {x−5=0, x+3=0}&{x−5=0, x+3=0}\\ {x=5, x=−3}&{x=5, x=−3}\\ {}&{(5,0) \text{and} (−3,0)}\\ {}&{\text{x-intercepts}}\\ \end{array}\]

    The solutions of the quadratic equation 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 equations 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 equation of the form \(ax^2+bx+c=0\). Now, we can use the discriminant to tell us how many x-intercepts there are on the graph.

    This figure shows three graphs side by side. The leftmost graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the lower right quadrant. Below the graph is the inequality b squared minus 4 a c greater than 0. Below that is the statement “Two solutions”. Below that is the statement “ Two x-intercepts”. The middle graph shows an downward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is on the x-axis. Below the graph is the equation b squared minus 4 a c equals 0. Below that is the statement “One solution”. Below that is the statement “ One x-intercept”. The rightmost graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the upper left quadrant. Below the graph is the inequality b squared minus 4 a c less than 0. Below that is the statement “No real solutions”. Below that is the statement “ No x-intercept”.

    Before you start solving the quadratic equation to find the values of the x-intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

    Example \(\PageIndex{13}\)

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

    Answer
      .
    To find the y-intercept, let x=0 and solve for y. .
    .
    When x=0, then y=4.
    The y-intercept is the point (0,4).
      .
    To find the x-intercept, let y=0 and solve for x. .
    Find the value of the discriminant to predict the number of solutions and so x-intercepts.

    b^2−4ac

    1^2−4⋅5⋅4

    1−80

    −79

    Since the value of the discriminant is negative, there is no real solution to the equation. There are no x-intercepts.

    Example \(\PageIndex{14}\)

    Find the intercepts of the parabola \(y=3x^2+4x+4\).

    Answer

    y:(0,4); x:none

    Example \(\PageIndex{15}\)

    Find the intercepts of the parabola \(y=x^2−4x−5\).

    Answer

    y:(0,−5); x:(5,0)(−1,0)

    Example \(\PageIndex{16}\)

    Find the intercepts of the parabola \(y=4x^2−12x+9\).

    Answer
      .
    To find the y-intercept, let x=0 and solve for y. .
    .
      When x=0, then y=9.
    The y-intercept is the point (0,9).
      .
    To find the x-intercept, let y=0 and solve for x. .
    Find the value of the discriminant to predict the number of solutions and so x-intercepts.

    b^2−4ac

    12^2−4⋅4⋅9

    144−144

    0

      Since the value of the discriminant is 0, there is no real solution to the equation. So there is one x-intercept.
    Solve the equation by factoring the perfect square trinomial. .
    Use the Zero Product Property. .
    Solve for x. .
    .
      When y=0, then \(\frac{3}{2}\)=x.
      The x-intercept is the point \((\frac{3}{2},0)\).

    Example \(\PageIndex{17}\)

    Find the intercepts of the parabola \(y=−x^2−12x−36.\).

    Answer

    y:(0,−36); x:(−6,0)

    Example \(\PageIndex{18}\)

    Find the intercepts of the parabola \(y=9x^2+12x+4\).

    Answer

    y:(0,4); x:\((−\frac{2}{3},0)\)

    Graph Quadratic Equations in Two Variables

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

    How To Graph a Quadratic Equation in Two Variables

    Example \(\PageIndex{19}\)

    Graph \(y=x2−6x+8\).

    Answer

    The image shows the steps to graph the quadratic equation y equals x squared minus 6 x plus 8. Step 1 is to write the quadratic equation with y on one side. This equation has y on one side already. The value of a is one, the value of b is -6 and the value of c is 8.Step 2 is to determine whether the parabola opens upward or downward. Since a is positive, the parabola opens upward.Step 3 is to find the axis of symmetry. The axis of symmetry is the line x equals negative b divided by the quantity 2 a. Plugging in the values of b and a the formula becomes x equals negative -6 divided by the quantity 2 times 1 which simplifies to x equals 3. The axis of symmetry is the line x equals 3.Step 4 is to find the vertex. The vertex is on the axis of symmetry. Substitute x equals 3 into the equation and solve for y. The equation is y equals x squared minus 6 x plus 8. Replacing x with 3 it becomes y equals 3 squared minus 6 times 3 plus 8 which simplifies to y equals -1. The vertex is (3, -1).Step 5 is to find the y-intercept and find the point symmetric to the y-intercept across the axis of symmetry. We substitute x equals 0 into the equation. The equation is y equals x squared minus 6 x plus 8. Replacing x with 0 it becomes y equals 0 squared minus 6 times 0 plus 8 which simplifies to y equals 8. The y-intercept is (0, 8). 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 equals 3. A point 3 units to the right of the axis of symmetry has x equals 6. The point symmetric to the y-intercept is (6, 8).Step 6 is to find the x-intercepts. We substitute y equals 0 into the equation. The equation becomes 0 equals x squared minus 6 x plus 8. We can solve this quadratic equation by factoring to get 0 equals the quantity x minus 2 times the quantity x minus 4. Solve each equation to get x equals 2 and x equals 4. The x-intercepts are (2, 0) and (4, 0).Step 7 is to graph the parabola. We graph the vertex, intercepts, and the point symmetric to the y-intercept. We connect these five points to sketch the parabola. The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -2 to 10. The y-axis of the plane runs from -3 to 10. The vertex is at the point (3, -1). Four points are plotted on the curve at (0, 8), (6, 8), (2, 0) and (4, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 3.

    Example \(\PageIndex{20}\)

    Graph the parabola \(y=x^2+2x−8\).

    Answer

    y:(0,−8); x:(2,0),(−4,0);
    axis: x=−1; vertex: (−1,−9);

    The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The vertex is at the point (-1, -9). Three points are plotted on the curve at (0, -8), (2, 0) and (-4, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -1.
     

    Example \(\PageIndex{21}\)

    Graph the parabola \(y=x^2−8x+12\).

    Answer

    y:(0,12); x:(2,0),(6,0);
    axis: x=4; vertex:(4,−4);

    The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The vertex is at the point (4, -4). Three points are plotted on the curve at (0, 12), (2, 0) and (6, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 4.
     

    Definition: GRAPH A QUADRATIC EQUATION IN TWO VARIABLES.

    1. Write the quadratic equation with yy on one side.
    2. Determine whether the parabola opens upward or downward.
    3. Find the axis of symmetry.
    4. Find the vertex.
    5. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
    6. Find the x-intercepts.
    7. 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 \(\PageIndex{22}\)

    Graph \(y=−x^2+6x−9\).

    Answer
    The equation y has on one side. .
    Since a is −1, the parabola opens downward.

    To find the axis of symmetry, find \(x=−\frac{b}{2a}\).
    . .
    .
    .

    The axis of symmetry is x=3. The vertex is on the line x=3.

    .

    Find y when x=3.
    .
    .
    .
    .
    The vertex is (3,0).

    .
    The y-intercept occurs when x=0.
    Substitute x=0.
    Simplify.

    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)
    .
    .
    .
    (0,−9).
    .
    The x-intercept occurs when y=0. .
    Substitute y=0. .
    Factor the GCF. .
    Factor the trinomial. .
    Solve for x. .
    Connect the points to graph the parabola. .
     

    Example \(\PageIndex{23}\)

    Graph the parabola \(y=−3x^2+12x−12\).

    Answer

    y:(0,−12); x:(2,0);
    axis: x=2; vertex:(2,0);

    The graph shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -1 to 10. The vertex is at the point (2, 0). One other point is plotted on the curve at (0, -12). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2.

    Example \(\PageIndex{24}\)

    Graph the parabola \(y=25x^2+10x+1\).

    Answer

    y:(0,1); x:(−15,0);
    axis: x=−15; vertex:(−15,0);

    The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -5 to 5. The y-axis of the plane runs from -5 to 10. The vertex is at the point (-1 fifth, 0). One other point is plotted on the curve at (0, 1). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -1 fifth.

    For the graph of \(y=−x^2+6x−9\) 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 \(y=x^2+4x+5\)?

    Example \(\PageIndex{25}\)

    Graph \(y=x^2+4x+5\).

    Answer
    The equation has y on one side. .
    Since a is 1, the parabola opens upward. .
    \(x=−\frac{b}{2a}\). .
    .
    .
    x=−2.
    .
    The vertex is on the line x=−2.  
    Find y when x=−2. .
    .
    .
    .
    (−2,1).
    .
    The y-intercept occurs when x=0.
    Substitute x=0.
    Simplify.
    The point (0,5) is two units to the right of the line of symmetry.
    The point two units to the left of the line of symmetry is (−4,5).
    .
    .
    .
    (0,5).
    .
    (−4,5)
    The x- intercept occurs when y=0.
    Substitute y=0.
    Test the discriminant.
    .
    .
        \(b^2−4ac\)
    \(42−4⋅15\)
    \(16−20\)
    \(−4\)
    Since the value of the discriminant is negative, there is no solution and so no x- intercept.
    Connect the points to graph the parabola. You may want to choose two more points for greater accuracy.
    .

    Example \(\PageIndex{26}\)

    Graph the parabola \(y=2x^2−6x+5\).

    Answer

    y:(0,5); x:none;
    axis: \(x=\frac{3}{2}\); vertex:\((\frac{3}{2},\frac{1}{2})\);

    The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -5 to 5. The y-axis of the plane runs from -5 to 10. The vertex is at the point (3 halves, 1 half). One other point is plotted on the curve at (0, 5). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 3 halves.

    Example \(\PageIndex{27}\)

    Graph the parabola \(y=−2x^2−1\).

    Answer

    y:(0,−1); x:none;
    axis: x=0; vertex:(0,−1);

    The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The vertex is at the point (0, -1). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 0.

    Finding the y-intercept by substituting x=0 into the equation is easy, isn’t it? But we needed to use the Quadratic Formula to find the x-intercepts in Example. We will use the Quadratic Formula again in the next example.

    Example \(\PageIndex{28}\)

    Graph \(y=2x^2−4x−3\).

    Answer
      .
    The equation y has one side.
    Since a is 2, the parabola opens upward.
    .
    To find the axis of symmetry, find\(x=−\frac{b}{2a}\) .
    .
    .
    The vertex is x=1
    The vertex on the line x=1. .
    Find y when x=1 .
    .
    .
    (1,−5)
    The y-intercept occurs when x=0. .
    Substitute x=0. .
    Simplify. .
    The y-intercept is (0,−3)

    The point (0,−3) is one unit to the left of the line of symmetry.
    The point one unit to the right of the line of symmetry is (2,−3)
    Point symmetric to the y-intercept is (2,−3).
    The x-intercept occurs when y=0 .
    Substitute y=0 .
    Use the Quadratic Formula. .
    Substitute in the values of a, b, c. .
    Simplify. .
    Simplify inside the radical. .
    Simplify the radical. .
    Factor the GCF. .
    Remove common factors. .
    Write as two equations. .
    Approximate the values. .
     
    The approximate values of the x-intercepts are (2.5,0) and (−0.6,0).
    Graph the parabola using the points found. .

    Example \(\PageIndex{29}\)

    Graph the parabola \(y=5x^2+10x+3\).

    Answer

    y:(0,3); x:(−1.6,0),(−0.4,0);
    axis: x=−1; vertex:(−1,−2);

    The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -5 to 5. The y-axis of the plane runs from -5 to 5. The vertex is at the point (-1,-2). Three other points are plotted on the curve at (0, 3), (-1.6, 0), (-0.4, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -1.

    Example \(\PageIndex{30}\)

    Graph the parabola \(y=−3x^2−6x+5\).

    Answer

    y:(0,5); x:(0.6,0),(−2.6,0);
    axis: x=−1; vertex:(−1,8);

    The graph shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The vertex is at the point (-1, 8). Three other points are plotted on the curve at (0, 5), (0.6, 0) and (-2.6, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -1.

    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 equation. The y-coordinate of the vertex is the minimum y-value of a parabola that opens upward. It is the maximum y-value of a parabola that opens downward. See Figure.

    This figure shows two graphs side by side. The left graph shows an downward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the upper right quadrant. The vertex is labeled “maximum”. The right graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the lower right quadrant. The vertex is labeled “minimum”.

    Definition: MINIMUM OR MAXIMUM VALUES OF A QUADRATIC EQUATION

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

    • minimum value of the quadratic equation if the parabola opens upward.
    • maximum value of the quadratic equation if the parabola opens downward.

    Example \(\PageIndex{31}\)

    Find the minimum value of the quadratic equation \(y=x^2+2x−8\).

    Answer
      .
    Since a is positive, the parabola opens upward.  
    The quadratic equation has a minimum.  
    Find the axis of symmetry. .
    .
    .
    x=−1
    The vertex is on the line x=−1. .
    Find y when x=−1. .
    .
    .
    (−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. .

    Example \(\PageIndex{32}\)

    Find the maximum or minimum value of the quadratic equation \(y=x^2−8x+12\).

    Answer

    The minimum value is −4 when x=4.

    Example \(\PageIndex{33}\)

    Find the maximum or minimum value of the quadratic equation \(y=−4x^2+16x−11\).

    Answer

    The maximum value is 5 when x=2.

    We have used the formula

    \[\begin{array} {l} {h=−16t^2+v_{0}t+h_{0}}\\ \nonumber \end{array}\]

    to calculate the height in feet, h, of an object shot upwards into the air with initial velocity, \(v_{0}\), after t seconds.

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

    Example \(\PageIndex{34}\)

    The quadratic equation \(h=−16t^2+v_{0}t+h_{0}\) models the height of a volleyball hit straight upwards with velocity 176 feet per second from a height of 4 feet.

    1. How many seconds will it take the volleyball to reach its maximum height?
    2. Find the maximum height of the volleyball.
    Answer

    \(h=−16t^2+176t+4\)

    Since a is negative, the parabola opens downward.

    The quadratic equation has a maximum.

    1.
    \[\begin{array} {ll} {}&{t=−\frac{b}{2a}}\\ {\text{Find the axis of symmetry.}}& {t=−\frac{176}{2(−16)}}\\ {}&{t=5.5}\\ {}&{\text{The axis of symmetry is} t = 5.5}\\ {\text{The vertex is on the line} t=5.5}& {\text{The maximum occurs when} t =5.5 \text{seconds.}}\\ \nonumber \end{array}\]

    2.

    Find h when t=5.5. .
    .
    Use a calculator to simplify. .
      The vertex is (5.5,488)
    Since the parabola has a maximum, the h-coordinate of the vertex is the maximum y-value of the quadratic equation. The maximum value of the quadratic is 488 feet and it occurs when t=5.5 seconds.

    Example \(\PageIndex{35}\)

    The quadratic equation \(h=−16t^2+128t+32\) 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? Round answers to the nearest tenth.

    Answer

    It will take 4 seconds to reach the maximum height of 288 feet.

    Example \(\PageIndex{36}\)

    A toy rocket shot upward from the ground at a rate of 208 ft/sec has the quadratic equation of \(h=−16t^2+208t\). When will the rocket reach its maximum height? What will be the maximum height? Round answers to the nearest tenth.

    Answer

    It will take 6.5 seconds to reach the maximum height of 676 feet.

    Access these online resources for additional instruction and practice graphing quadratic equations:

    Key Concepts

    • The graph of every quadratic equation is a parabola.
    • Parabola Orientation For the quadratic equation \(y=ax^2+bx+c\), if
      • a>0, the parabola opens upward.
      • a<0, the parabola opens downward.
    • Axis of Symmetry and Vertex of a Parabola For a parabola with equation \(y=ax^2+bx+c\):
      • The axis of symmetry of a parabola is the line \(x=−\frac{b}{2a}\).
      • The vertex is on the axis of symmetry, so its x-coordinate is \(−\frac{b}{2a}\).
      • To find the y-coordinate of the vertex we substitute \(x=−\frac{b}{2a}\) into the quadratic equation.
    • Find the Intercepts of a Parabola To find the intercepts of a parabola with equation \(y=ax^2+bx+c\):
      \[\begin{array} {ll} {\textbf{y-intercept}}&{\textbf{x-intercepts}}\\ {\text{Let} x=0 \text{and solve for y}}&{\text{Let} y=0 \text{and solve for x}}\\ \nonumber \end{array}\]
    • To Graph a Quadratic Equation in Two Variables
      1. Write the quadratic equation with yy on one side.
      2. Determine whether the parabola opens upward or downward.
      3. Find the axis of symmetry.
      4. Find the vertex.
      5. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
      6. Find the x-intercepts.
      7. Graph the parabola.
    • Minimum or Maximum Values of a Quadratic Equation
      • The y-coordinate of the vertex of the graph of a quadratic equation is the
      • minimum value of the quadratic equation if the parabola opens upward.
      • maximum value of the quadratic equation if the parabola opens downward.

     

    Glossary

    axis of symmetry
    The axis of symmetry is the vertical line passing through the middle of the parabola \(y=ax^2+bx+c\).
    parabola
    The graph of a quadratic equation in two variables is a parabola.
    quadratic equation in two variables
    A quadratic equation in two variables, where a, b, and c are real numbers and \(a \ge 0\) is an equation of the form \(y=ax^2+bx+c\).
    vertex
    The point on the parabola that is on the axis of symmetry is called the vertex of the parabola; it is the lowest or highest point on the parabola, depending on whether the parabola opens upwards or downwards.
    x-intercepts of a parabola
    The x-intercepts are the points on the parabola where \(y=0\).
    y-intercept of a parabola
    The y-intercept is the point on the parabola where \(x=0\).