Quadratic Functions and Modeling

Graphing Quadratic Functions

Transformations of Quadratic Functions

Quadratic functions can be graphed by transforming the parent function f(x)=x2f(x) = x^2.
A parent function is a function of a certain family type that has the simplest algebraic rule. The quadratic parent function is:
f(x)=x2f(x)=x^2
The graph of the parent function is a parabola that opens upward and has its vertex at the origin. The yy-axis is the axis of symmetry, and the parabola passes through the points (1,1)(1, 1) and (1,1)(-1, 1).

A transformation of a graph is a change in its position or shape. One way to graph quadratic functions is to transform the graph of the parent function. It can be transformed by horizontal and vertical translations, vertical stretches and compressions, and vertical reflections.

A horizontal translation is a shift to the left or to the right. For example, a shift of the parent function of a quadratic function hh units to the left is represented by:
f(x)=(x+h)2f(x)=(x+h)^{2}

Horizontal Translations

Horizontal translations of the parent function of a quadratic function are shifts hh units to the left or to the right. For instance, to graph f(x)=(x+7)2f(x)=(x+7)^2, translate the parent graph 7 units to the left. To graph f(x)=(x5)2f(x)=(x-5)^2, translate the parent graph 5 units to the right.
A vertical translation is a shift up or down. For example, a shift of the parent function of a quadratic function kk units down is represented by:
f(x)=x2kf(x)=x^{2}-k

Vertical Translations

Vertical translations are shifts of the parent function either up or down. To graph f(x)=x2+6f(x)=x^2+6, translate the parent graph 6 units up. To graph f(x)=x24f(x)=x^2-4, translate the parent graph 4 units down.
A parabola can be made narrower or wider using vertical stretches and compressions. A vertical stretch pulls the points of a graph away from the xx-axis, and a vertical compression pushes the points of a graph toward the xx-axis. A parabola can also be reflected across an axis, resulting in a mirror image of itself.

Vertical Stretches, Compressions, and Reflections

Vertical stretches, compressions, or reflections depend on the slope of the function. To graph f(x)=3x2f(x)=3x^2, vertically stretch the parent graph by a factor of 3. To graph f(x)=18x2f(x)=\frac{1}{8}x^2, vertically compress the parent graph by a factor of 18\frac{1}{8}. To graph f(x)=x2f(x)=-x^2, reflect the parent graph across the xx-axis.
The vertex form of a quadratic function is an equation, where aa, hh, and kk are real numbers and a0a\neq0, written as:
f(x)=a(xh)2+kf(x)=a(x-h)^2+k
It is called the vertex form because it makes it easier to find the vertex of the parabola, (h,k)(h, k). The values hh and kk are used to determine the horizontal and vertical translations. The value of aa is used to determine the vertical stretch or compression. If the absolute value of aa is greater than 1, the result is a vertical stretch. If the absolute value of aa is less than 1, the result is a vertical compression.
Step-By-Step Example
Graphing a Quadratic Function in Vertex Form
Graph the given function:
f(x)=13(x+2)2+1f(x) = -\frac{1}{3}(x + 2)^2 + 1
Step 1
The vertex form of a quadratic function is:
f(x)=a(xh)2+kf(x)=a(x-h)^2+k
Rewrite the given function using the vertex form of a quadratic function:
f(x)=13(x+2)2+1f(x)=13(x(2))2+1\begin{aligned}f(x) &= -\frac{1}{3}(x + 2)^2 + 1\\f(x) &= -\frac{1}{3}(x -(-2))^2 + 1\end{aligned}
Step 2

Use the value of a|a| to identify a vertical compression or vertical stretch.

  • If 0<a<10 < |a| < 1, the parent graph is compressed vertically.
  • If a>1|a| > 1, the parent graph is stretched vertically.
f(x)=13(x(2))2+1a=13a=13<1\begin{aligned}f(x) &= -\frac{1}{3}(x -(-2))^2 + 1\\a&=-\frac{1}{3}\\|a| &= \frac{1}{3}<1\end{aligned}
Then graph the compression of the parent graph by a factor of 13\frac{1}{3}.
Step 3
Look at the sign of aa. If a<0a < 0, the graph from Step 2 is reflected across the xx-axis.
f(x)=13(x(2))2+1a=13<0\begin{aligned}f(x) &= -\frac{1}{3}(x -(-2))^2 + 1\\a &= -\frac{1}{3}<0\end{aligned}
The value of aa is negative. So, graph the compressed graph from Step 2 to show a reflection across the xx-axis.
Solution
Identify the vertex (h,k)({\color{#c42126}{h}}, {\color{#0047af}{k}}) in the vertex form of the given function:
f(x)=13(x(2))2+1f(x)=-\frac{1}{3}(x -({\color{#c42126}{-2}}))^2 + {\color{#0047af}{1}}
The values for hh and kk are h=2h =-2; and k=1k = 1. So, the vertex is at (2,1)(-2, 1). Translate the graph from Step 3 so that the vertex is at the point (2,1)(-2, 1).

Deriving the Coordinates of the Vertex

The values of aa, bb, and cc can be used to determine the vertex and axis of symmetry of f(x)=ax2+bx+cf(x) = ax^2 + bx + c.
The standard form of a quadratic function is a quadratic equation, where aa, bb, and cc are real numbers and a0a\neq0, written in the form:
f(x)=ax2+bx+cf(x)=ax^2+bx+c
To determine the coordinates of the vertex, compare the standard form with the vertex form of the quadratic function. Start by expanding the vertex form. Then set the xx-terms from the two forms equal to each other and solve for hh, the xx-coordinate of the vertex.
The xx-coordinate of the vertex of the standard form of a quadratic equation is the value of hh, or b2a-\frac{b}{2a}.
The result is:
b2a-\frac{b}{2a}
Next, set the constant terms from the two forms equal to each other and solve for kk, the yy-coordinate of the vertex.
The yy-coordinate of the vertex of the standard form of a quadratic equation is the value of kk, or cb24ac-\frac{b^2}{4a}.
The result is:
cb24ac-\frac{b^2}{4a}
Therefore, the coordinates of the vertex are:
(b2a,cb24a)\left ( -\frac{b}{2a}, c-\frac{b^2}{4a} \right )
The axis of symmetry passes through the vertex, so the equation of the axis of symmetry is:
x=b2ax = -\frac{b}{2a}

Graphing a Quadratic Function in Standard Form

A quadratic function of the form f(x)=ax2+bx+cf(x) = ax^2 + bx + c can be graphed by locating the vertex and the axis of symmetry.
The standard form of a quadratic function is:
f(x)=ax2+bx+cf(x) = ax^2 + bx + c
To graph the standard form, use the values of aa, bb, and cc to identify and plot the vertex and the axis of symmetry. Then plot at least one other point that is not the vertex, and sketch a parabola through the points.
Step-By-Step Example
Using Standard Form to Graph a Quadratic Function
Graph the function:
f(x)=2x28x+1f(x) = 2x^2 - 8x + 1
Step 1
The standard form of a quadratic function is:
f(x)=ax2+bx+cf(x) = ax^2 + bx + c
Identify the values of aa, bb, and cc in the given function.
f(x)=2x28x+1f(x) = 2x^2 - 8x + 1
The values for aa, bb, and cc are:
a=2b=8c=1\begin{aligned}a &= 2\\b&=-8\\c &= 1\end{aligned}
Notice that bb is negative.
Step 2

Use the values of aa, bb, and cc to find the vertex and axis of symmetry.

  • Tip: For the formula for the yy-coordinate of the vertex, substitute the xx-coordinate of the vertex into the function.
  • Use the xx-coordinate of the vertex for the axis of symmetry.
Vertex:
(b2a,cb24a)=(822,1(8)242)=(84,1648)(h,k)=(2,7)\begin{aligned}\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)&=\left(-\frac{-8}{2\cdot 2},1-\frac{(-8)^2}{4 \cdot 2}\right)\\&=\left(\frac{8}{4},1-\frac{64}{8}\right)\\(h,k) &=(2,-7)\end{aligned}
Axis of symmetry: x=2x = 2
Solution

Plot the vertex, and draw the axis of symmetry. Then plot at least one other point on the graph:

  • The yy-intercept is f(0)=cf(0) = c. Since c=1c=1, the yy-intercept is 1.
  • From the vertex, move 1 unit right and aa units up (if a>0a>0) or down (if a<0a<0) to locate another point. Since a=2a=2, move 1 unit right and 2 units up from the vertex.
  • Use symmetry to locate additional points.
Sketch the parabola through the points.