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11.4: Graph quadratic functions

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    45113
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    Let’s recall the parabola from the Functions chapter.

    Example 11.4.1

    Graph \(f(x)=x^2\).

    Solution

    Let’s pick five \(x\)-coordinates, and find corresponding \(y\)-values. Each \(x\)-value being positive, negative, and zero. This is common practice, but not required.

    Table 11.4.1
    \(x\) \(f(x)=x^2\) \((x,f(x))\)
    \(-2\) \(f(\color{blue}{-2}\color{black}{)}=4\) \((-2,4)\)
    \(-1\) \(f(\color{blue}{-1}\color{black}{)}=1\) \((-1,1)\)
    \(0\) \(f(\color{blue}{0}\color{black}{)}=0\) \((0,0)\)
    \(1\) \(f(\color{blue}{1}\color{black}{)}=1\) \((1,1)\)
    \(2\) \(f(\color{blue}{2}\color{black}{)}=4\) \((2,4)\)

    Plot the five ordered-pairs from the table. To connect the points, be sure to connect them from smallest \(x\)-value to largest \(x\)-value, i.e., left to right. This graph is called a parabola and since this function is quite common for the \(x^2\)-form, we call it a quadratic (square) function.

    clipboard_eb5c12ee1d6759dfb55be1c3c325f92ad.png
    Figure 11.4.1

    Since quadratic functions have a leading term that contains \(x^2\), then a quadratic function’s graph is called a parabola just like in the Functions chapter.

    Definition: Quadratic Function

    A quadratic function is a polynomial function of the form

    \[f(x)=ax^2+bx+c\nonumber\]

    where \(a\neq 0\).

    In Example 11.4.1 , we plotted points and connected the dots. This is one way of graphing quadratic functions, but not the most efficient. Hence, we can easily graph quadratic functions by finding key elements of the function: vertex, \(x\)-intercepts, and the \(y\)-intercept.

    Vertex of a Quadratic Function

    Definition: Vertex

    The vertex of a quadratic function \(f(x)=ax^2+bx+x\) is given by

    \[\left(-\dfrac{b}{2a},f\left(-\dfrac{b}{2a}\right)\right)\nonumber\]

    Example 11.4.2

    Find the vertex of \(f(x) = x^2 − 3x − 4\).

    Solution

    The \(x\)-coordinate of the vertex is \(−\dfrac{b}{2a}\) given by the definition. In this case, \(a = 1\), \(b = −3\), and \(c = −4\). Hence,

    \[x=-\dfrac{b}{2a}=-\dfrac{-3}{2(1)}=\dfrac{3}{2}\nonumber\]

    The \(y\)-coordinate of the vertex is

    \[f\left(\dfrac{3}{2}\right)=\left(\dfrac{3}{2}\right)^2-3\left(\dfrac{3}{2}\right)-4=-\dfrac{25}{4}\nonumber\]

    Thus, the vertex of \(f(x)\) is at \(\left(\dfrac{3}{2},-\dfrac{25}{4}\right)\). Let’s take a look at the graph and verify this is the location of the vertex. We see the vertex is, in fact, located at \(\left(\dfrac{3}{2},-\dfrac{25}{4}\right)\). Additionally, we see the parabola intersects the \(x\)-axis at \(x = −1\) and \(x = 4\), and the \(y\)-axis at \((0, −4)\).

    clipboard_e9e63ea9d53f4d24f54298e25fd0e8408.png
    Figure 11.4.2

    Graph Quadratic Functions by its Properties

    Properties of a Quadratic Function

    To graph a quadratic function, \(f(x) = ax^2 + bx + x\), by its properties, we obtain key properties.

    Property 1. The direction of the parabola.

    • If \(a > 0\), then the graph is an upward parabola.
    • If \(a < 0\), then the graph is a downward parabola.

    Property 2. The vertex: \(\left(-\dfrac{b}{2a},f\left(-\dfrac{b}{2a}\right)\right)\)

    Property 3. The \(y\)-intercept: \((0, f(0))\)

    Property 4. The \(x\)-intercepts: \((x, 0)\), i.e., where \(f(x) = 0\), also known as the zeros of \(f(x)\).

    Property 5. The axis of symmetry: \(x = −\dfrac{b}{2a}\)

    Note

    The axis of symmetry is a vertical line that intersects the vertex of the parabola. Hence, the line \(x=-\dfrac{b}{2a}\). The axis of symmetry essentially “cuts” the parabola in half and the parabola is symmetrical about this axis.

    Example 11.4.3

    Using the properties, graph \(f(x)=x^2+4x+3\).

    Solution

    Property 1. The direction of the parabola. Since \(a = 1\) and \(a > 0\), then \(f(x)\) is an upwards parabola.

    Property 2. Find the vertex. We use the formula to find the vertex, where \(a = 1\) and \(b = 4\). \[\begin{array}{rl}x=-\dfrac{b}{2a}&\text{Plug-n-chug} \\ x=-\dfrac{\color{blue}{4}}{\color{blue}{(2)(1)}}&\color{black}{\text{Simplify}} \\ x=-2&\text{The }x\text{-coordinate of the vertex}\end{array}\nonumber\] Next, we find the \(y\)-coordinate of the vertex by obtaining \(f(-2)\). \[\begin{array}{rl}f(x)=x^2+4x+3&\text{Plug-n-chug} \\ f(\color{blue}{-2}\color{black}{)}=(\color{blue}{-2}\color{black}{)}^2+4(\color{blue}{-2}\color{black}{)}+3&\text{Evaluate} \\ f(-2)=-1&\text{The }y\text{-coordinate of the vertex}\end{array}\nonumber\] Hence, the vertex is \((-2,-1)\).

    Property 3. Find the \(y\)-intercept. We can find the \(y\)-intercept by obtaining \(f(0)\). \[\begin{array}{rl}f(x)=x^2+4x+3&\text{Plug-n-chug }x=0 \\ f(\color{blue}{0}\color{black}{)}=\color{blue}{0}\color{black}{^2}+4(\color{blue}{0}\color{black}{)}+3 &\text{Evaluate} \\ f(0)=3&\text{The }y\text{-intercept}\end{array}\nonumber\] Hence, the \(y\)-intercept is \((0,3)\).

    Property 4. Find the \(x\)-intercepts. We can find the \(x\)-intercept by obtaining where \(f(x) = 0\). \[\begin{array}{rl} f(x)=x^2+4x+3&\text{Plug-n-chug }f(x)=0 \\ 0=x^2+4x+3&\text{Factor} \\ 0=(x+3)(x+1)&\text{Apply the zero product rule} \\ x+3=0\quad\text{and}\quad x+1=0&\text{Solve each equation} \\ x=-3\quad\text{and}\quad x=-1&\text{The }x\text{-intercepts}\end{array}\nonumber\] Hence, the \(x\)-intercepts are \((-3,0)\) and \((-1,0)\).

    Property 5. Find the axis of symmetry. We see from the vertex in Property 2. \(x = −2\). Thus, the axis of symmetry is the vertical line \(x = −2\).

    Let’s graph all the properties and label each property.

    clipboard_e7fc6cd74b2346336d6738c2e67c6b5c7.png
    Figure 11.4.3

    So, we verify that \(f(x)\) is an upwards parabola with \(x\)-intercepts \((−3, 0)\) and \((−1, 0)\), \(y\)-intercept \((0, 3)\), vertex \((−2, −1)\), and axis of symmetry \(x = −2\).

    Example 11.4.4

    Using the properties, graph \(g(x) = −3x^2 + 12x − 9\).

    Solution

    Property 1. The direction of the parabola. Since \(a = −3\) and \(a < 0\), then \(g(x)\) is an downwards parabola.

    Property 2. Find the vertex. We use the formula to find the vertex, where \(a = −3\) and \(b = 12\). \[\begin{array}{rl}x=-\dfrac{b}{2a}&\text{Plug-n-chug} \\ x=-\dfrac{\color{blue}{12}}{\color{blue}{2(-3)}}&\text{Simplify} \\ x=2&\text{The }x\text{-coordinate of the vertex}\end{array}\nonumber\] Next, we find the \(y\)-coordinate of the vertex by obtaining \(g(2)\). \[\begin{array}{rl}g(x)=-3x^2+12x-9&\text{Plug-n-chug} \\ g(\color{blue}{2}\color{black}{)}=-3(\color{blue}{2}\color{black}{)}^2+12(\color{blue}{2}\color{black}{)}-9&\text{Evaluate} \\ g(2)=3&\text{The }y\text{-coordinate of the vertex}\end{array}\nonumber\] Hence, the vertex is \((2,3)\).

    Property 3. Find the \(y\)-intercept. We can find the \(y\)-intercept by obtaining \(f(0)\). \[\begin{array}{rl}g(x)=-3x^2+12x-9&\text{ Plug-n-chug }x=0 \\ g(\color{blue}{0}\color{black}{)}=-3\color{blue}{0}\color{black}{^2}+12(\color{blue}{0}\color{black}{)}-9&\text{Evaluate} \\ g(0)=-9&\text{The }y\text{-intercept}\end{array}\nonumber\] Hence, the \(y\)-intercept is \((0,-9)\).

    Property 4. Find the \(x\)-intercepts. We can find the \(x\)-intercept by obtaining where \(g(x) = 0\). \[\begin{array}{rl} g(x)=-3x^2+12x-9&\text{Plug-n-chug }g(x)=0 \\ 0=-3x^2+12x-9&\text{Factor a GCF of }-3 \\ =-3(x^2-4x+3)&\text{Divide each side by }-3\text{, then factor} \\ 0=(x-3)(x-1)&\text{Apply the zero product rule} \\ x-3=0\quad\text{and}\quad x-1=0&\text{Solve each equation} \\ x=3\quad\text{and}\quad x=1&\text{The }x\text{-intercepts}\end{array}\nonumber\] Hence, the \(x\)-intercepts are \((3,0)\) and \((1,0)\).

    Property 5. Find the axis of symmetry. We see from the vertex in Property 2. \(x = 2\). Thus, the axis of symmetry is the vertical line \(x = 2\).

    Let’s graph all the properties and label each property.

    clipboard_e5de861e990ef370e7556c953977ed9d6.png
    Figure 11.4.4

    So, we verify that \(g(x)\) is an downwards parabola with \(x\)-intercepts \((3, 0)\) and \((1, 0)\), \(y\)-intercept \((0, −9)\), vertex \((2, 3)\), and axis of symmetry \(x = 2\).

    Graph Quadratic Functions by Transformations

    Definition: Vertex Form

    A quadratic function in vertex form is given as

    \[f(x)=a(x-h)^2+k,\nonumber\]

    where the domain consists of all real numbers and \((h, k)\) is the vertex.

    Recall. In the chapter with rational functions, we graphed rational functions with horizontal and vertical shift. Let’s take this one step further, and graph quadratic functions with not only horizontal and vertical shifts, but also with a stretch or compression.

    Transformations of Quadratic Functions

    Given \(f(x)\) is the quadratic function in vertex form

    \[f(x)=a(x-h)^2+k,\nonumber\]

    horizontal and vertical shifts, and vertical stretches or compressions of \(f(x)\) are described below:

    Table 11.4.1
    \(f(x-h)\) \(f(x)\pm k\) \(af(x)\)
    Transformation Horizontal shift Vertical shift Vertical stretch or compression
    Units \(h>0\): Shift \(h\) units to the right
    \(h<0\): Shift \(h\) units to the left
    \(k>0\): Shift \(k\) units upwards
    \(k<0\): Shift \(k\) units downwards
    \(|a|>1\): Multiply all outputs by \(a\); vertical stretch
    \(0<|a|<1\): Multiply all outputs by \(a\); vertical compression

    Recall from the previous subsection, if \(a > 0\), the parabola is upward and if \(a < 0\), the parabola is downward.

    Example 11.4.5

    Using the library function \(f(x)=x^2\), graph \(g(x)=x^2+2\).

    Solution

    We start by noticing we are adding \(2\) to \(f(x)\), i.e., \(g(x) = f(x) + 2\):

    \[\begin{aligned}g(x)&=x^2+2 \\ g(x)&=f(x)+2\end{aligned}\]

    This means, from the table, \(g(x)\) has a vertical shift by \(2\) units upward. Let’s start with \(f(x) = x^2\), and then shift \(f(x)\) \(2\) units upward to obtain \(g(x)\):

    clipboard_ef133507eddaf00de76a6ae2fca727c42.png
    Figure 11.4.5

    We can see that the blue solid graph is \(g(x)\), where \(g(x)\) is an upward parabola, the axis of symmetry is \(x = 0\). Notice, all points on \(f(x)\) shifted upwards by \(2\) units. We can use the well-defined points of the library function \(f(x) = x^2\) to transform into \(g(x)\).

    Example 11.4.6

    Using the library function \(f(x) = x^2\), graph \(h(x) = (x − 3)^2 + 2\).

    Solution

    We start by noticing we are adding \(2\) to \(f(x)\) and subtracting \(3\) from the input \(x\), i.e., \(h(x) = f(x − 3) + 2\):

    \[\begin{aligned}h(x)&=(x-3)^2+2 \\ h(x)&=f(x-3)+2\end{aligned}\]

    This means, from the table, \(h(x)\) has a horizontal shift to the right by \(3\) units, and vertical shift by \(2\) units upward. Let’s start with \(f(x) = x^2\), and then shift \(f(x)\) \(3\) units to the right, and \(2\) units upward to obtain \(h(x)\):

    clipboard_ed84bb2018f63f21c0b3771c8393f12c1.png
    Figure 11.4.6

    We can see that the blue solid graph is \(h(x)\), where \(h(x)\) is an upward parabola, the axis of symmetry is \(x = 3\). Notice, all points on \(f(x)\) shifted right by \(3\) units and upwards by \(2\) units. We can use the well-defined points of the library function \(f(x) = x^2\) to transform into \(h(x)\).

    Example 11.4.7

    Using the library function \(f(x) = x^2\), graph \(k(x) = −2(x + 1)^2 − 3\).

    Solution

    We start by noticing we are subtracting \(3\) from \(f(x)\), vertically stretching by a factor of \(−2\), and adding \(1\) to the input \(x\), i.e., \(k(x) = −2\cdot f(x + 1) − 3\):

    \[\begin{aligned} k(x)&=-2(x+1)^2-3 \\ k(x)&=-2\cdot f(x+1)-3\end{aligned}\]

    This means, from the table, \(k(x)\) has a horizontal shift to the left by \(1\) unit, a vertical stretch by a factor of \(−2\), and a vertical shift by \(3\) units downward. Let’s start with \(f(x) = x^2\), and then apply these transformations to obtain \(k(x)\). Since there are three transformations, it is best we split this up into three steps.

    Step 1. Graph the library function, \(f(x)=x^2\), and apply the horizontal shift: \(f(x+1)\).

    clipboard_ef6b56ce9977a07281866d94422aed0cb.png
    Figure 11.4.7

    Step 2. Graph \(f(x + 1)\) from Step 1., and apply the vertical stretch, i.e., multiply the \(y\)-coordinates of the well-defined ordered pairs by \(−2\): \(−2\cdot f(x + 1)\)

    clipboard_ea333c7a8aa992c1e2cd962a59dce8bc5.png
    Figure 11.4.8

    Step 3. Graph \(−2\cdot f(x + 1)\) from Step 2., and apply the vertical shift: \(−2\cdot f(x + 1) − 3\)

    clipboard_ebeee74c4b80425040107afb0655e2a49.png
    Figure 11.4.9

    We can see that the blue solid graph is \(k(x)\), where \(k(x)\) is an downward parabola, the axis of symmetry is \(x = −1\). Notice, all points on \(f(x)\) shifted left by \(1\) unit, stretched by a factor of \(−2\), and shifted downwards by \(3\) units.

    Note

    Notice with three transformations, it can get tedious and tricky. The best route to apply multiple transformations is to follow order of operations, e.g., first parenthesis, multiplication/division, and then addition/subtraction. With transformations, this translates to

    Step 1. Apply the horizontal shift

    Step 2. Vertically stretch or compress the function

    Step 3. Lastly, apply the vertical shift

    A way to remember the order in which we apply the transformations is \(hak\): first the \(h\), then \(a\), lastly, \(k\).

    Graph Quadratic Functions Homework

    Graph the quadratic function using the properties. Be sure to label your graph with all properties.

    Exercise 11.4.1

    \(f(x)=x^2-2x-8\)

    Exercise 11.4.2

    \(f(x)=2x^2-12x+10\)

    Exercise 11.4.3

    \(f(x)=-2x^2+12x-18\)

    Exercise 11.4.4

    \(f(x)=-3x^2+24x-45\)

    Exercise 11.4.5

    \(f(x)=-x^2+4x+5\)

    Exercise 11.4.6

    \(f(x)=-x^2+6x-5\)

    Exercise 11.4.7

    \(f(x)=-2x^2+16x-24\)

    Exercise 11.4.8

    \(f(x)=3x^2+12x+9\)

    Exercise 11.4.9

    \(f(x)=5x^2-40x+75\)

    Exercise 11.4.10

    \(f(x)=-5x^2-60x-175\)

    Exercise 11.4.11

    \(f(x)=x^2-2x-3\)

    Exercise 11.4.12

    \(f(x)=2x^2-12x+16\)

    Exercise 11.4.13

    \(f(x)=-2x^2+12x-10\)

    Exercise 11.4.14

    \(f(x)=-3x^2+12x-9\)

    Exercise 11.4.15

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

    Exercise 11.4.16

    \(f(x)=-2x^2+16x-30\)

    Exercise 11.4.17

    \(f(x)=2x^2+4x-6\)

    Exercise 11.4.18

    \(f(x)=5x^2+30x+45\)

    Exercise 11.4.19

    \(f(x)=5x^2+20x+15\)

    Exercise 11.4.20

    \(f(x)=-5x^2+20x-15\)

    Starting with the library function \(y=x^2\), state the function, \(f(x)\), given its transformation(s).

    Exercise 11.4.21

    vertically stretched by a factor of \(3\) and shifted right \(1\)

    Exercise 11.4.22

    vertically stretched by a factor of \(-2\), and shifted left \(3\)

    Exercise 11.4.23

    vertically compressed by a factor of \(\dfrac{1}{3}\)

    Exercise 11.4.24

    vertically stretched by a factor of \(2\) and shifted right \(4\)

    Exercise 11.4.25

    vertically stretched by a factor of \(4\), shifted left \(4\)

    Exercise 11.4.26

    vertically compressed by a factor of \(−\dfrac{1}{2}\) and shifted upward by \(3\) units

    Exercise 11.4.27

    vertically stretched by a factor of \(3\) and shifted down \(4\)

    Exercise 11.4.28

    vertically stretched by a factor of \(2\), shifted right \(3\), and shifted up \(1\)

    Starting with the library function \(y=x^2\), graph the function using transformations.

    Exercise 11.4.29

    \(g(x)=-\dfrac{1}{2}(x+7)^2\)

    Exercise 11.4.30

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

    Exercise 11.4.31

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

    Exercise 11.4.32

    \(f(x)=\dfrac{1}{2}(x+2)^2+9\)

    Exercise 11.4.33

    \(f(x)=2(x+4)^2-5\)

    Exercise 11.4.34

    \(f(x)=-2(x-4)^2+7\)

    Exercise 11.4.35

    \(g(x)=2(x-3)^2-2\)

    Exercise 11.4.36

    \(g(x)=-\dfrac{1}{2}(x+5)^2\)

    Exercise 11.4.37

    \(f(x)=\dfrac{1}{2}(x+4)^2+8\)

    Exercise 11.4.38

    \(f(x)=-2(x-8)^2+7\)


    This page titled 11.4: Graph quadratic functions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Darlene Diaz (ASCCC Open Educational Resources Initiative) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.