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5.1.2: A.2- Graphing Quadratic Equations Using Transformations

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    94091
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    Learning Objectives

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

    • Graph quadratic expressions of the form \(y=x^{2}+k\)
    • Graph quadratic expressions of the form \(y=(x−h)^{2}\)
    • Graph quadratic expressions of the form \(y=ax^{2}\)
    • Graph quadratic expressions using transformations
    • Find a quadratic expressions from its graph
    Be Prepared

    Before you get started, take this readiness quiz.

    1. Graph the expressions \(y=x^{2}\) by plotting points.
    2. Factor completely: \(y^{2}−14y+49\).
    3. Factor completely: \(2x^{2}−16x+32\).

    Graphing Quadratic Expressions of the Form \(y=x^{2}+k\)

    In the last section, we learned how to graph quadratic expressions using their properties. Another method involves starting with the basic graph of \(y=x^{2}\) and ‘moving’ it according to information given in the equation. We call this graphing quadratic equations using transformations.

    In the first example, we will graph the quadratic equation \(y=x^{2}\) by plotting points. Then we will see what effect adding a constant, \(k\), to the equation will have on the graph of the new equation \(y=x^{2}+k\).

    Example \(\PageIndex{1}\)

    Graph \(y=x^{2}\), \(y=x^{2}+2\), and \(y=x^{2}−2\) on the same rectangular coordinate system. Describe what effect adding a constant to the right side of the equation has on the basic parabola.

    Solution:

    Plotting points will help us see the effect of the constants on the basic \(y=x^{2}\) graph. We fill in the chart for all three equations.

    A table depicting the effect of constants on the basic equation y equals x squared. The table has seven columns labeled x, y equals x squared, the ordered pair (x, y), y equals x squared plus 2, the ordered pair (x, y), y equals x squared minus 2, and the ordered pair (x, y). In the x column, the values given are negative 3, negative 2, negative 1, 0, 1, 2, and 3. In the y equals x squared column, the values are 9, 4, 1, 0, 1, 4, and 9. In the (x, y) column, the ordered pairs (negative 3, 9), (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), (2, 4), and (3, 9) are given. The y equals x squared plus 2 column contains the expressions 9 plus 2, 4 plus 2, 1 plus 2, 0 plus 2, 1 plus 2, 4 plus 2, and 9 plus 2. The (x, y) column has the ordered pairs of (negative 3, 11), (negative 2, 6), (negative 1, 3), (0, 2), (1, 3), (2, 6), and (3, 11). In the y equals x squared minus 2 column, the expressions given are 9 minus 2, 4 minus 2, 1 minus 2, 0 minus 2, 1 minus 2, 4 minus 2, and 9 minus 2. In last column, (x, y), contains the ordered pairs (negative 3, 7), (negative 2, 2), (negative 1, negative 1), (0, negative 2), (1, negative 1), (2, 2), and (3, 7).

    The \(y\)-coordinates of the solutions to \(y=x^2+2\) are two more than the \(y\)-coordinates of the solutions to \(y=x^2\) values. Also, the \(y\)-coordinates of the solutions to \(y=x^2-2\) values are two less than the \(y\)-coordinates of the solutions to \(y=x^2\) values. Now we will graph all three equations on the same rectangular coordinate system.

    This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle is the graph of y equals x squared has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The top parabola has been moved up 2 units, and the bottom has been moved down 2 units.


    The graph of \(y=x^{2}+2\) is the same as the graph of \(y=x^{2}\) but shifted up \(2\) units.

    The graph of \(y=x^{2}−2\) is the same as the graph of \(y=x^{2}\) but shifted down \(2\) units.

    Exercise \(\PageIndex{1}\)
    1. Graph \(y=x^{2}, y=x^{2}+1,\) and \(y=x^{2}-1\) on the same rectangular coordinate system.
    2. Describe what effect adding a constant to the equation has on the basic parabola.
    Answer

    a.

    This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle graph is of y equals x squared has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The top curve has been moved up 1 unit, and the bottom has been moved down 1 unit.This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle graph is of y equals x squared has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The top curve has been moved up 1 unit, and the bottom has been moved down 1 unit.
    Figure 9.7.3

    b. The graph of \(y=x^{2}+1\) is the same as the graph of \(y=x^{2}\) but shifted up \(1\) unit. The graph of \(y=x^{2}−1\) is the same as the graph of \(y=x^{2}\) but shifted down \(1\) unit.

    Exercise \(\PageIndex{2}\)
    1. Graph \(y=x^{2}, y=x^{2}+6,\) and \(y=x^{2}-6\) on the same rectangular coordinate system.
    2. Describe what effect adding a constant to the equation has on the basic parabola.
    Answer

    a.

    This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle curve is the graph of y equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The top curve has been moved up 6 units, and the bottom has been moved down 6 units.
    Figure 9.7.4

    b. The graph of \(y=x^{2}+6\) is the same as the graph of \(y=x^{2}\) but shifted up \(6\) units. The graph of \(y=x^{2}-6\) is the same as the graph of \(y=x^{2}\) but shifted down \(6\) units.

    The last example shows us that to graph a quadratic equation of the form \(y=x^{2}+k\), we take the basic parabola graph of \(y=x^{2}\) and vertically shift it up \((k>0)\) or shift it down \((k<0)\).

    This transformation is called a vertical shift.

    Graphing a Quadratic equation of the Form \(y=x^{2}+k\) Using a Vertical Shift

    The graph of \(y=x^{2}+k\) shifts the graph of \(y=x^{2}\) vertically \(k\) units.

    • If \(k>0\), shift the parabola vertically up \(k\) units.
    • If \(k<0\), shift the parabola vertically down \(|k|\) units.

    Now that we have seen the effect of the constant, \(k\), it is easy to graph equations of the form \(y=x^{2}+k\). We just start with the basic parabola of \(y=x^{2}\) and then shift it up or down.

    It may be helpful to practice sketching \(y=x^{2}\) quickly. We know the values and can sketch the graph from there.

    This figure shows an upward-opening parabola on the x y-coordinate plane, with vertex (0, 0). Other points on the curve are located at (negative 4, 16), (negative 3, 9), (negative 2, 4), (negative 1, 1), (1, 1), (2, 4), (3, 9), and (4, 16).
    Figure 9.7.5

    Once we know this parabola, it will be easy to apply the transformations. The next example will require a vertical shift.

    Example \(\PageIndex{2}\)

    Graph \(y=x^{2}−3\) using a vertical shift.

    Solution:

    We first draw the graph of \(y=x^{2}\) on the grid. This figure shows an upward-opening parabola on the x y-coordinate plane with a vertex of (0, 0) with other points on the curve located at (negative 1, 1) and (1, 1). It is the graph of y equals x squared.
    Determine \(k\).

    \(y=x^2+k\)

    \(y=x^2-3\)

      .
    Shift the graph \(y=x^{2}\) down \(3\). CNX_IntAlg_Figure_09_07_004d_img1.jpg
    Table 9.7.1
    Exercise \(\PageIndex{3}\)

    Graph \(y=x^{2}−5\) using a vertical shift.

    Answer

    This figure shows 2 upward-opening parabolas on the x y-coordinate plane. The top curve is the graph of y equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The bottom curve has been moved down 5 units.
    Exercise \(\PageIndex{4}\)

    Graph \(y=x^{2}+7\) using a vertical shift.

    Answer

    This figure shows 2 upward-opening parabolas on the x y-coordinate plane. The bottom curve is the graph of y equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The top curve has been moved up 7 units.

    Graphing Quadratic equations of the Form \(y=(x-h)^{2}\)

    In the first example, we graphed the quadratic equation \(y=x^{2}\) by plotting points and then saw the effect of adding a constant \(k\) to the right side of the equation (or, equivalently, subtracting \(k\) from \(y\)) had on the resulting graph of the new equation \(y=x^{2}+k\).

    We will now explore the effect of subtracting a constant, \(h\), from \(x\) has on the resulting graph of the new equation \(y=(x−h)^{2}\).

    Example \(\PageIndex{3}\)

    Graph \(y=x^{2}, y=(x-1)^{2},\) and \(y=(x+1)^{2}\) on the same rectangular coordinate system. Describe what effect adding a constant to the equation has on the basic parabola.

    Solution:

    Plotting points will help us see the effect of the constants on the basic \(y=x^{2}\) graph. We fill in the chart for all three equations.


    A table depicting the effect of constants on the basic function of x squared. The table has seven columns labeled x, f of x equals x squared, the ordered pair (x, f of x), g of x equals the quantity of x minus 1 squared, the ordered pair (x, g of x), h of x equals the quantity of x plus 1 squared, and the ordered pair (x, h of x). In the x column, the values given are negative 3, negative 2, negative 1, 0, 1, 2, and 3. In the f of x equals x squared column, the values are 9, 4, 1, 0, 1, 4, and 9. In the (x, f of x) column, the ordered pairs (negative 3, 9), (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), (2, 4), and (3, 9) are given. The g of x equals the quantity of x minus 1 squared column contains the values of 16, 9, 4, 1, 0, 1, and 4. The (x, g of x) column has the ordered pairs of (negative 3, 1), (negative 2, 9), (negative 1, 4), (0, 1), (1, 0), (2, 1), and (3, 4). In the h of x equals the quantity of x plus 1 squared, the values given are 4, 1, 0, 1, 4, 9, and 16. In last column, (x, h of x), contains the ordered pairs (negative 3, 4), (negative 2, 1), (negative 1, 0), (0, 4), (1, negative 1), (2, 9), and (3, 16).
    Figure 9.7.12

    The \(y\)-coordinates share the common numbers \(0, 1, 4, 9\), and \(16\), but are shifted, in that they correspond to different \(x\)-coordinates in the different cases.

    This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle curve is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The left curve has been moved to the left 1 unit, and the right curve has been moved to the right 1 unit.
    The figure says on the first line that the graph of g of x equals the quantity x minus 1 square is the same as the graph of f of x equals x squared but shifted right 1 unit. The second line states that the graph of h of x equals the quantity x plus 1 squared is the same as the graph of f of x equals x squared but shifted left 1 unit. The third line of the figure says g of x equals the quantity x minus 1 squared with an arrow underneath it pointing to the right with 1 unit written beside it. Finally, it gives h of x equals the quantity of x plus 1 squared with an arrow underneath it pointing to the left with 1 unit written beside it.
    Exercise \(\PageIndex{5}\)
    1. Graph \(y=x^{2}, y=(x+2)^{2},\) and \(y=(x-2)^{2}\) on the same rectangular coordinate system.
    2. Describe what effect adding a constant to the \(x\) variable has on the basic parabola.
    Answer

    a.

    This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle curve is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The left curve has been moved to the left 2 units, and the right curve has been moved to the right 2 units.

    b. The graph of \(y=(x+2)^{2}\) is the same as the graph of \(y=x^{2}\) but shifted left \(2\) units. The graph of \(y=(x−2)^{2}\) is the same as the graph of \(y=x^{2}\) but shifted right \(2\) units.

    Exercise \(\PageIndex{6}\)
    1. Graph \(y=x^{2}, y=x^{2}+5,\) and \(y=x^{2}-5\) on the same rectangular coordinate system.
    2. Describe what effect adding a constant to the equation has on the basic parabola.
    Answer

    a.

    This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle curve is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The left curve has been moved to the left 5 units, and the right curve has been moved to the right 5 units.

    b. The graph of \(y=(x+5)^{2}\) is the same as the graph of \(y=x^{2}\) but shifted left \(5\) units. The graph of \(y=(x-5)^{2}\) is the same as the graph of \(y=x^{2}\) but shifted right \(5\) units.

    The last example shows us that to graph a quadratic equation of the form \(y=(x−h)^{2}\), we take the basic parabola graph of \(y=x^{2}\) and shift it left \((h>0)\) or shift it right \((h<0)\).

    This transformation is called a horizontal shift.

    Graphing a Quadratic Equation of the Form \(y=(x-h)^{2}\) Using a Horizontal Shift

    The graph of \(y=(x-h)^{2}\) shifts the graph of \(y=x^{2}\) horizontally \(h\) units.

    • If \(h>0\), shift the parabola horizontally left \(h\) units.
    • If \(h<0\), shift the parabola horizontally right \(|h|\) units.

    Now that we have seen the effect of the constant, \(h\), it is easy to graph equations of the form \(y=(x−h)^{2}\). We just start with the basic parabola of \(y=x^{2}\) and then shift it left or right.

    The next example will require a horizontal shift.

    Example \(\PageIndex{4}\)

    Graph \(y=(x−5)^{2}\) using a horizontal shift.

    Solution:

    We first draw the graph of \(y=x^{2}\) on the grid. .
    Determine \(h\).

    \(y=(x-h)^2\)

    \(y=(x-5)^2\)

      \(h=5\)
    Shift the graph \(y=x^{2}\) to the right \(5\) units.

    CNX_IntAlg_Figure_09_07_008d_img1.jpg

    Table 9.7.2
    Exercise \(\PageIndex{7}\)

    Graph \(y=(x−4)^{2}\) using a horizontal shift.

    Answer
    This figure shows 2 upward-opening parabolas on the x y-coordinate plane. The left curve is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The right curve has been moved right 4 units.
    Exercise \(\PageIndex{8}\)

    Graph \(y=(x+6)^{2}\) using a horizontal shift.

    Answer
    This figure shows 2 upward-opening parabolas on the x y-coordinate plane. The right curve is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The left curve has been moved to the left 6 units.

    (Note that you may also view the consideration of the equation \(y=x^2+k\) as \(y-k=x^2\) and think of this as a shift along the \(y\)-axis. Viewing it this way allows for a similar approach to the transformation of graphs of equations formed by adding (or subtracting) constants to (or from) variables.)

    Now that we know the effect of the constants \(h\) and \(k\), we will graph a quadratic equation of the form \(y=(x-h)^{2}+k\) by first drawing the basic parabola and then making a horizontal shift followed by a vertical shift. We could do the vertical shift followed by the horizontal shift, but most students prefer the horizontal shift followed by the vertical.

    Example \(\PageIndex{5}\)

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

    Solution:

    This equation will involve two transformations and we need a plan.

    Let’s first identify the constants \(h, k\).

    \(y=(x+1)^2-2\)

    \(y=(x-h)^2+k\)

    \(y=(x-(-1))^2-2\)

    \(h=-1, k=-2\)

    The \(h\) constant gives us a horizontal shift and the \(k\) gives us a vertical shift.

    y equals x squared is given with an arrow coming from it pointing to y equals the quantity x plus 1 squared with an arrow coming from it pointing to f of x equals the quantity x plus 1 squared minus 2. The next lines say h equals negative 1 which means shift left 1 unit and k equals negative 2 which means shift down 2 units.

    We first draw the graph of \(y=x^{2}\) on the grid.

    The figure says on the first line that the graph of y equals the quantity x plus 1 squared is the same as the graph of y equals x squared but shifted left 1 unit. The second line states that the graph of y equals the quantity x plus 1 squared minus 2 is the same as the graph of y equals the quantity x plus 1 squared but shifted down 2 units.
    The first graph shows 1 upward-opening parabola on the x y-coordinate plane. It is the graph of y equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). By shifting that graph of y equals x squared left 1, we move to the next graph, which shows the original y equals x squared and then another curve moved left one unit to produce y equals the quantity of x plus 1 squared. By moving f of x equals the quantity of x plus 1 squared down 1, we move to the final graph, which shows the original f of x equals x squared and the f of x equals the quantity of x plus 1, then another curve moved down 1 to produce y equals the quantity of x plus 1 squared minus 2.
    Exercise \(\PageIndex{9}\)

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

    Answer
    This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of y equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). Then, the original function is moved 2 units to the left to produce y equals the quantity of x plus 2 squared. The final curve is produced by moving down 3 units to produce y equals the quantity of x plus 2 squared minus 3.
    Exercise \(\PageIndex{10}\)

    Graph \(y=(x-3)^{2}+1\) using transformations.

    Answer
    This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of y equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). Then, the original function is moved 3 units to the right to produce y equals the quantity of x minus 3 squared. The final curve is produced by moving up 1 unit to produce y equals the quantity of x minus 3squared plus 1.

    Graphing Quadratic Equations of the Form \(y=ax^{2}\)

    So far we graphed the quadratic equation \(y=x^{2}\) and then saw the effect of including a constant \(h\) or \(k\) in the equation had on the resulting graph of the new equation. We will now explore the effect of the coefficient \(a\) on the resulting graph of the new equation \(y=ax^{2}\).

    A table depicting the effect of constants on the graph of y equals squared. The table has seven columns labeled x, y equals x squared, the ordered pair (x, y), y equals 2 times x squared, the ordered pair (x, y), y equals one-half times x squared, and the ordered pair (x, y). In the x column, the values given are negative 2, negative 1, 0, 1, and 2. In the y equals x squared column, the values are 4, 1, 0, 1, and 4. In the (x, y) column, the ordered pairs (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), and (2, 4) are given. The y equals 2 times x squared column contains the expressions 2 times 4, 2 times 1, 2 times 0, 2 times 1, and 2 times 4. The (x, y) column has the ordered pairs of (negative 2, 8), (negative 1, 2), (0, 0), (1, 2), and (2,8). In the y equals one-half times x squared, the expressions given are one-half times 4, one-half times 1, one-half times 0, one-half times 1, and one-half times 4. In last column, (x, y), contains the ordered pairs (negative 2, 2), (negative 1, one-half), (0, 0), (1, one-half), and (2, 2).

    If we graph these equations, we can see the effect of the constant \(a\), assuming \(a>0\).

    This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of y equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The slimmer curve of y equals 2 times x square has a vertex at (0,0) and other points of (negative 1, one-half) and (1, one-half). The wider curve, y equals one-half x squared, has a vertex at (0,0) and other points of (negative 2, 2) and (2,2).

    The graph of the equation \(y=2x^2\) is "skinnier" than the graph of \(y=x^2\).

    The graph of the equation \(y=\dfrac12 x^2\) is "wider" than the graph of \(y=x^2\).

    To graph a equation with constant \(a\) it is easiest to choose a few points on \(y=x^{2}\) and multiply the \(y\)-coordinates by \(a\).

    Graphing a Quadratic equation of the Form \(y=ax^{2}\)

    The coefficient \(a\) in the equation \(y=ax^{2}\) affects the graph of \(y=x^{2}\) by stretching or compressing it.

    • If \(0<|a|<1\), the graph of \(y=ax^{2}\) will be “wider” than the graph of \(y=x^{2}\).
    • If \(|a|>1\), the graph of \(y=ax^{2}\) will be “skinnier” than the graph of \(y=x^{2}\).
    Example \(\PageIndex{6}\)

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

    Solution:

    We will graph the equations \(y=x^{2}\) and \(y=3x^{2}\) on the same grid. We will choose a few points on \(y=x^{2}\) and then multiply the \(y\)-values by \(3\) to get the points for \(y=3x^{2}\).

    The table depicts the effect of constants on the basic graph of y equals x squared. The table has 3 columns labeled x, y equals x squared with the ordered pair (x, y), and y equals 3 times x squared with the ordered pair (x, y). In the x column, the values given are negative 2, negative 1, 0, 1, and 2. In the y equals x squared with the ordered pair (x, y), the ordered pairs (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), and (2, 4) are given. The y equals 3 times x squared with the ordered pair (x, y) column has the ordered pairs of (negative 2, 12) because 3 times 4 equals 12, (negative 1, 3) because 3 times 1 equals 3, (0, 0) because 3 times 0 equals 0, (1, 3) because 3 times 1 equals 3, and (2,12) because 3 times 4 equals 12. The graph beside the table shows 2 upward-opening parabolas on the x y-coordinate plane. One is the graph of y equals x squared and has a vertex of (0, 0). Other points given on the curve are located at (negative 2, 4) (negative 1, 1), (1, 1), and (2,4). The slimmer curve of y equals 3 times x squared has a vertex at (0,0) and other points given of (negative 2, 12), (negative 1, 3), (1, 3), and (2,12).
    Exercise \(\PageIndex{11}\)

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

    Answer
    The graph shows the upward-opening parabola on the x y-coordinate plane of y equals x squared that has a vertex of (0, 0). Other points given on the curve are located at (negative 2, 4) (negative 1, 1), (1, 1), and (2,4). Also shown is a downward-opening parabola of y equals negative 3 times x squared. It has a vertex of (0,0) with other points at (negative 1, negative 3) and (1, negative 3)
    Exercise \(\PageIndex{12}\)

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

    Answer
    This figure shows 2 upward-opening parabolas on the x y-coordinate plane. One is the graph of y equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The slimmer curve of y equals 2 times x square has a vertex at (0,0) and other points of (negative 1, one-half) and (1, one-half).

    Graphing Quadratic Equations of the Form \(y=ax^{2}+bx+c\) Using Transformations

    We have learned how the constants \(a, h\), and \(k\) in the equations, \(y=x^{2}+k, y=(x−h)^{2}\), and \(y=ax^{2}\) affect their graphs. We can now put this together and graph quadratic equations \(y=ax^{2}+bx+c\) by first putting them into the form \(y=a(x−h)^{2}+k\) by completing the square. This form is sometimes known as the vertex form or standard form.

    We must be careful to both add and subtract the number to the expression to complete the square. We cannot add the number to 'both sides' (both sides of what?) as we did when we completed the square with quadratic equations. Note that we could also add and subtract a number from the same side in the case of the quadratic equation as well.

    Quadratic Equation Quadratic Expression
    \(x^2+8x+6=0\) \(x^2+8x+6\)
    \(x^2+8x=-6\) \(x^2+8x+6\)
    \(x^2+8x+16=-6+16\) -- add \(16\) to both sides \(x^2+8x+16-16+6\) --add and subtract \(16\) from the expression
    \((x+4)^2=10\) \((x+4)^2-10\)

    When we complete the square in a equation with a coefficient of \(x^{2}\) that is not one, we have to factor that coefficient from just the \(x\)-terms. We do not factor it from the constant term. It is often helpful to move the constant term a bit to the right to make it easier to focus only on the \(x\)-terms.

    Once we get the constant we want to complete the square, we must remember to multiply it by that coefficient before we then subtract it.

    Example \(\PageIndex{7}\)

    Rewrite \(y=−3x^{2}−6x−1\) in the \(y=a(x−h)^{2}+k\) form by completing the square.

    Solution:

     

    \(y=-3x^2-6x-1\)

    Separate the \(x\) terms from the constant.

    \(y=-3x^2-6x\quad-1\)

    Factor the coefficient of \(x^{2}, -3\).

    \(y=-3(x^2+2x)\quad-1\)

    Prepare to complete the square.

    \(y=-3(x^2+2x\quad\quad)\quad-1\)

    Take half of \(2\) and then square it to complete the square \((\frac{1}{2}\cdot 2)^{2}=1\)  
    The constant \(1\) completes the square in the parentheses, but the parentheses is multiplied by \(-3\). So we are really adding \(-3\). We must then add \(3\) to not change the value of the equation.

    \(y=-3((x^2+2x+1-1))-1\)

    \(y=-3(x^2+2x+1)+3-1\)

    Rewrite the trinomial as a square and subtract the constants.

    \(y=-3(x+1)^2+2\)

    The equation is now in the \(y=a(x-h)^{2}+k\) form.

    \(y=a(x-h)^2+k\)

    \(y=-3(x+1)^2+2\)

    Exercise \(\PageIndex{13}\)

    Rewrite \(y=−4x^{2}−8x+1\) in the \(y=a(x−h)^{2}+k\) form by completing the square.

    Answer

    \(y=-4(x+1)^{2}+5\)

    Exercise \(\PageIndex{14}\)

    Rewrite \(y=2x^{2}−8x+3\) in the \(y=a(x−h)^{2}+k\) form by completing the square.

    Answer

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

    Once we put the equation into the \(y=(x−h)^{2}+k\) form, we can then use the transformations as we did in the last few problems. The next example will show us how to do this.

    Example \(\PageIndex{8}\)

    Graph \(y=x^{2}+6x+5\) by using transformations.

    Solution:

    Step 1: Rewrite the equation in \(y=a(x-h)^{2}+k\) vertex form by completing the square.

     

    \(y=x^2+6x+5\)

    Separate the \(x\) terms from the constant. \(y=x^2+6x+5\)
    Take half of \(6\) and then square it to complete the square. \((\frac{1}{2}\cdot 6)^{2}=9\)  
    We both add \(9\) and subtract \(9\) to not change the value of the equation.

    \(y=x^2+6x+9-9+5\)

    Rewrite the trinomial as a square and subtract the constants.

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

    The equation is now in the \(y=(x-h)^{2}+k\) form.

    \(y=(x-h)^2+k\)

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

    Step 2: Graph the equation using transformations.

    Looking at the \(h, k\) values, we see the graph will take the graph of \(y=x^{2}\) and shift it to the left \(3\) units and down \(4\) units.

    y equals x squared is given with an arrow coming from it pointing to y equals the quantity x plus 3 squared with an arrow coming from it pointing to y equals the quantity x plus 3 squared minus 4. The next lines say h equals negative 3 which means shift left 3 unit and k equals negative 4 which means shift down 4 units

    We first draw the graph of \(y=x^{2}\) on the grid.

    To graph y equals the quantity x plus 3 squared, shift the graph of y equals x squares to the left 3 units. To graph y equals the quantity x plus 3 squared minus 4, shift the graph the quantity x plus 3 squared down 4 units.
    The first graph shows 1 upward-opening parabola on the x y-coordinate plane. It is the graph of y equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). By shifting that graph of y equals x squared left 3, we move to the next graph, which shows the original y equals x squared and then another curve moved left 3 units to produce y equals the quantity of x plus 3 squared. By moving y equals the quantity of x plus 3 squared down 2, we move to the final graph, which shows the original y equals x squared and the y equals the quantity of x plus 3 squared, then another curve moved down 4 to produce y equals the quantity of x plus 1 squared minus 4.
    Exercise \(\PageIndex{15}\)

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

    Answer
    This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of y equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The curve to the left has been moved 1 unit to the left to produce y equals the quantity of x plus 1 squared. The third graph has been moved down 4 units to produce y equals the quantity of x plus 1 squared minus 4.
    Exercise \(\PageIndex{16}\)

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

    Answer
    This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of y equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The curve to the right has been moved 4 units to the right to produce y equals the quantity of x minus 4 squared. The third graph has been moved down 4 units to produce y equals the quantity of x minus 4 squared minus 4.

    We list the steps to take a graph a quadratic equation using transformations here.

    Graphing a Quadratic Equation of the Form \(y=ax^{2}+bx+c\) Using Transformations

    1. Rewrite the equation in \(y=a(x-h)^{2}+k\) form by completing the square.
    2. Graph the equation using transformations.
    Example \(\PageIndex{9}\)

    Graph \(y=-2x^{2}-4x+2\) by using transformations.

    Solution:

    Step 1: Rewrite the equation in \(y=a(x-h)^{2}+k\) vertex form by completing the square.

     

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

    Separate the \(x\) terms from the constant.

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

    We need the coefficient of \(x^{2}\) to be one. We factor \(-2\) from the \(x\)-terms.

    \(y=-2(x^2+2x\quad)+2\)

    Take half of \(2\) and then square it to complete the square. \((\frac{1}{2}\cdot 2)^{2}=1\)  
    We add \(1\) to complete the square in the parentheses, but the parentheses is multiplied by \(-2\). So we are really adding \(-2\). To not change the value of the equation we add \(2\).

    \(y=-2((x^2+2x+1)-1)+2\)

    \(y=-2(x^2+2x+1)+2+2\)

    Rewrite the trinomial as a square ad subtract the constants.

    \(y=-2(x+1)^2+4\)

    The equation is now in the \(y=a(x-h)^{2}+k\) form.

    \(y=a(x-h)^2+k\)

    \(y=-2(x+1)^2+4\)

    Step 2: Graph the equation using transformations.

    y equals x squared is given with an arrow coming from it pointing to y equals negative 2 times x squared with an arrow coming from it pointing to y equals negative 2 times the quantity x plus 1 squared. An arrow come from it to point to y equals negative 2 times the quantity x plus 1 squared plus 4. The next line says a equals negative 2 which means multiply the y-values by negative 2, then h equals negative 1 which means shift left 1 unit and k equals 4 which means shift up 4 units

    We first draw the graph of \(y=x^{2}\) on the grid.

    To graph y equals negative 2 times x squared, multiply the y-values in parabola of y equals x squared by negative 2. To graph y equals negative 2 times the quantity x plus 1 squared, shift the graph of y equals negative 2 times x squared to the left 1 unit. To graph y equals negative 2 times the quantity x plus 1 squared plus 4, shift the graph of y equals negative 2 times the quantity x plus 1 squared up 4 units.
    The first graph shows 1 upward-opening parabola on the x y-coordinate plane. It is the graph of y equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). By multiplying by negative 2, move to the next graph showing the original y equals x squared and the new slimmer and flipped graph of y equals negative 2 x squared. By shifting that graph of y equals negative 2 times x squared left 1, we move to the next graph, which shows the original y equals x squared, y equals negative 2 x squared, and then another curve moved left 1 unit to produce y equals negative 2 times the quantity of x plus 1 squared. By moving y equals negative 2 times the quantity of x plus 1 squared up 4, we move to the final graph, which shows the original y equals x squared, y equals negative 2 x squared, and the y equals negative 2 times the quantity of x plus 1 squared, then another curve moved up 4 to produce y equals negative 2 times the quantity of x plus 1 squared plus 4.
    Exercise \(\PageIndex{17}\)

    Graph \(y=-3x^{2}+12x-4\) by using transformations.

    Answer
    This figure shows a downward-opening parabola on the x y-coordinate plane with a vertex of (2,8) and other points of (1,5) and (3,5).
    Exercise \(\PageIndex{18}\)

    Graph \(y=−2x^{2}+12x−9\) by using transformations.

    Answer
    This figure shows a downward-opening parabola on the x y-coordinate plane with a vertex of (3, 9) and other points of (1, 1) and (5, 1).

    Now that we have completed the square to put a quadratic equation into \(y=a(x−h)^{2}+k\) form, we can also use this technique to graph the equation using its properties as in the previous section.

    If we look back at the last few examples, we see that the vertex is related to the constants \(h\) and \(k\).

    The first graph shows an upward-opening parabola on the x y-coordinate plane with a vertex of (negative 3, negative 4) with other points of (0, negative 5) and (0, negative 1). Underneath the graph, it shows the standard form of a parabola, y equals the quantity x minus h squared plus k, with the equation of the parabola y equals the quantity of x plus 3 squared minus 4 where h equals negative 3 and k equals negative 4. The second graph shows a downward-opening parabola on the x y-coordinate plane with a vertex of (negative 1, 4) and other points of (0,2) and (negative 2,2). Underneath the graph, it shows the standard form of a parabola, y equals a times the quantity x minus h squared plus k, with the equation of the parabola y equals negative 2 times the quantity of x plus 1 squared plus 4 where h equals negative 1 and k equals 4.

    In each case, the vertex is \((h,k)\). Also the axis of symmetry is the line \(x=h\).

    We rewrite our steps for graphing a quadratic equation using properties for when the equation is in \(y=a(x−h)^{2}+k\) form.

    Graphing a Quadratic Equation of the Form \(y=a(x-h)^{2}+k\) Using Properties

    1. Rewrite the equation \(y=a(x-h)^{2}+k\) form.
    2. Determine whether the parabola opens upward, \(a>0\), or downward, \(a<0\).
    3. Find the axis of symmetry, \(x=h\).
    4. Find the vertex, \((h,k)\).
    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.
    Example \(\PageIndex{10}\)
    1. Rewrite \(y=2 x^{2}+4 x+5\) in \(y=a(x-h)^{2}+k\) form
    2. Graph the equation using properties

    Solution:

    Rewrite the equation in \(y=a(x-h)^{2}+k\) form by completing the square. \(y=2 x^{2}+4 x+5\)
      \(y=2\left(x^{2}+2 x\right)+5\)
      \(y=2\left(x^{2}+2 x+1\right)+5-2\)
      \(y=2(x+1)^{2}+3\)
    Identify the constants \(a, h, k\).  
    Since \(a=2\), the parabola opens upward. .
    The axis of symmetry is \(x=h\). The axis of symmetry is \(x=-1\).
    The vertex is \((h,k)\). The vertex is \((-1,3)\).
    Find the \(y\)-intercept by finding \(f(0)\). \(f(0)=2 \cdot 0^{2}+4 \cdot 0+5\)
      \(f(0)=5\)
      \(y\)-intercept \((0,5)\)
    Find the point symmetric to \((0,5)\) across the axis of symmetry. \((-2,5)\)
    Find the \(x\)-intercepts. The discriminant is negative, so there are no \(x\)-intercepts. Graph the parabola.
      .
    Exercise \(\PageIndex{19}\)
    1. Rewrite \(y=3 x^{2}-6 x+5\) in \(y=a(x-h)^{2}+k\) form
    2. Graph the equation using properties
    Answer
    1. \(y=3(x-1)^{2}+2\)

    2. The graph shown is an upward facing parabola with vertex (1, 2) and y-intercept (0, 5). The axis of symmetry is shown, x equals 1.
      Figure 9.7.66
    Exercise \(\PageIndex{20}\)
    1. Rewrite \(y=-2 x^{2}+8 x-7\) in \(y=a(x-h)^{2}+k\) form
    2. Graph the equation using properties
    Answer
    1. \(y=-2(x-2)^{2}+1\)

    2. The graph shown is a downward facing parabola with vertex (2, 1) and x-intercepts (1, 0) and (3, 0). The axis of symmetry is shown, x equals 2.
      Figure 9.7.67

    Challenge section:

    Find a Quadratic Equation from its Graph

    So far we have started with a equation and then found its graph.

    Now we are going to reverse the process. Starting with the graph, we will find the equation.

    Example \(\PageIndex{11}\)

    Determine the quadratic equation whose graph is shown.

    The graph shown is an upward facing parabola with vertex (negative 2, negative 1) and y-intercept (0, 7).
    Figure 9.7.68

    Solution:

    Since it is quadratic, we start with the \(y=a(x−h)^{2}+k\) form.

    The vertex, \((h,k)\), is \((−2,−1)\) so \(h=−2\) and \(k=−1\).

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

    To find \(a\), we use the \(y\)-intercept, \((0,7)\).

    So \(f(0)=7\).

    \(7=a(0+2)^{2}-1\)

    Solve for \(a\).

    \(\begin{array}{l}{7=4 a-1} \\ {8=4 a} \\ {2=a}\end{array}\)

    Write the equation.

    \(y=a(x-h)^{2}+k\)

    Substitute in \(h=-2, k=-1\) and \(a=2\).

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

    Exercise \(\PageIndex{21}\)

    Write the quadratic equation in \(y=a(x−h)^{2}+k\) form whose graph is shown.

    The graph shown is an upward facing parabola with vertex (3, negative 4) and y-intercept (0, 5).
    Figure 9.7.69
    Answer

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

    Exercise \(\PageIndex{22}\)

    Determine the quadratic equation whose graph is shown.

    The graph shown is an upward facing parabola with vertex (negative 3, negative 1) and y-intercept (0, 8).
    Figure 9.7.70
    Answer

    \(y=(x+3)^{2}-1\)

    Key Concepts

    • Graph a Quadratic equation of the form \(y=x^{2}+k\) Using a Vertical Shift
      • The graph of \(y=x^{2}+k\) shifts the graph of \(y=x^{2}\) vertically \(k\) units.
        • If \(k>0\), shift the parabola vertically up \(k\) units.
        • If \(k<0\), shift the parabola vertically down \(|k|\) units.
    • Graph a Quadratic equation of the form \(y=(x−h)^{2}\) Using a Horizontal Shift
      • The graph of \(y=(x−h)^{2}\) shifts the graph of \(y=x^{2}\) horizontally \(h\) units.
        • If \(h>0\), shift the parabola horizontally left \(h\) units.
        • If \(h<0\), shift the parabola horizontally right \(|h|\) units.
    • Graph of a Quadratic equation of the form \(y=ax^{2}\)
      • The coefficient \(a\) in the equation \(y=ax^{2}\) affects the graph of \(y=x^{2}\) by stretching or compressing it.
        If \(0<|a|<1\), then the graph of \(y=ax^{2}\) will be “wider” than the graph of \(y=x^{2}\).
        If \(|a|>1\), then the graph of \(y=ax^{2}\) will be “skinnier” than the graph of \(y=x^{2}\).
    • How to graph a quadratic equation using transformations
      1. Rewrite the equation in \(y=a(x−h)^{2}+k\) form by completing the square.
      2. Graph the equation using transformations.
    • Graph a quadratic equation in the vertex form \(y=a(x−h)^{2}+k\) using properties
      1. Rewrite the equation in \(y=a(x−h)^{2}+k\) form.
      2. Determine whether the parabola opens upward, \(a>0\), or downward, \(a<0\).
      3. Find the axis of symmetry, \(x=h\).
      4. Find the vertex, \((h,k)\).
      5. Find the \(y\)-intercept. Find the point symmetric to the \(y\)-intercept across the axis of symmetry.
      6. Find the \(x\)-intercepts, if possible.
      7. Graph the parabola.

    5.1.2: A.2- Graphing Quadratic Equations Using Transformations is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.