### Transformations of Quadratic Functions

Quadratic functions can be graphed by transforming the parent function $f(x) = x^2$.

A
The graph of the parent function is a parabola that opens upward and has its vertex at the origin. The $y$-axis is the axis of symmetry, and the parabola passes through the points $(1, 1)$ and $(-1, 1)$.
A vertical translation is a shift up or down. For example, a shift of the parent function of a quadratic function $k$ units down is represented by:
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 $x$-axis, and a vertical compression pushes the points of a graph toward the $x$-axis. A parabola can also be reflected across an axis, resulting in a mirror image of itself.
The
It is called the vertex form because it makes it easier to find the vertex of the parabola, $(h, k)$. The values $h$ and $k$ are used to determine the horizontal and vertical translations. The value of $a$ is used to determine the vertical stretch or compression. If the absolute value of $a$ is greater than 1, the result is a vertical stretch. If the absolute value of $a$ is less than 1, the result is a vertical compression.

**parent function**is a function of a certain family type that has the simplest algebraic rule. The quadratic parent function is:$f(x)=x^2$

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 $h$ units to the left is represented by:$f(x)=(x+h)^{2}$

#### Horizontal Translations

$f(x)=x^{2}-k$

#### Vertical Translations

#### Vertical Stretches, Compressions, and Reflections

**vertex form of a quadratic function**is an equation, where $a$, $h$, and $k$ are real numbers and $a\neq0$, written as:$f(x)=a(x-h)^2+k$

Step-By-Step Example

Graphing a Quadratic Function in Vertex Form

Graph the given function:

$f(x) = -\frac{1}{3}(x + 2)^2 + 1$

Step 1

The vertex form of a quadratic function is:
Rewrite the given function using the vertex form of a quadratic function:

$f(x)=a(x-h)^2+k$

$\begin{aligned}f(x) &= -\frac{1}{3}(x + 2)^2 + 1\\f(x) &= -\frac{1}{3}(x -(-2))^2 + 1\end{aligned}$

Step 2

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

- If $0 < |a| < 1$, the parent graph is compressed vertically.
- If $|a| > 1$, the parent graph is stretched vertically.

$\begin{aligned}f(x) &= -\frac{1}{3}(x -(-2))^2 + 1\\a&=-\frac{1}{3}\\|a| &= \frac{1}{3}<1\end{aligned}$

Step 3

Look at the sign of $a$. If $a < 0$, the graph from Step 2 is reflected across the $x$-axis.
The value of $a$ is negative. So, graph the compressed graph from Step 2 to show a reflection across the $x$-axis.

$\begin{aligned}f(x) &= -\frac{1}{3}(x -(-2))^2 + 1\\a &= -\frac{1}{3}<0\end{aligned}$

Solution

Identify the vertex $({\color{#c42126}{h}}, {\color{#0047af}{k}})$ in the vertex form of the given function:
The values for $h$ and $k$ are $h =-2$; and $k = 1$. So, the vertex is at $(-2, 1)$.
Translate the graph from Step 3 so that the vertex is at the point $(-2, 1)$.

$f(x)=-\frac{1}{3}(x -({\color{#c42126}{-2}}))^2 + {\color{#0047af}{1}}$

### Deriving the Coordinates of the Vertex

The values of $a$, $b$, and $c$ can be used to determine the vertex and axis of symmetry of $f(x) = ax^2 + bx + c$.

The
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 $x$-terms from the two forms equal to each other and solve for $h$, the

The result is:
Next, set the constant terms from the two forms equal to each other and solve for $k$, the $y$-coordinate of the vertex.

The result is:
Therefore, the coordinates of the vertex are:
The axis of symmetry passes through the vertex, so the equation of the axis of symmetry is:

**standard form of a quadratic function**is a quadratic equation, where $a$, $b$, and $c$ are real numbers and $a\neq0$, written in the form:$f(x)=ax^2+bx+c$

*x*-coordinate of the vertex.The result is:

$-\frac{b}{2a}$

The result is:

$c-\frac{b^2}{4a}$

$\left ( -\frac{b}{2a}, c-\frac{b^2}{4a} \right )$

$x = -\frac{b}{2a}$

### Graphing a Quadratic Function in Standard Form

A quadratic function of the form $f(x) = ax^2 + bx + c$ can be graphed by locating the vertex and the axis of symmetry.

The standard form of a quadratic function is:
To graph the standard form, use the values of $a$, $b$, and $c$ 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.

$f(x) = ax^2 + bx + c$

Step-By-Step Example

Using Standard Form to Graph a Quadratic Function

Graph the function:

$f(x) = 2x^2 - 8x + 1$

Step 1

The standard form of a quadratic function is:
Identify the values of $a$, $b$, and $c$ in the given function.
The values for $a$, $b$, and $c$ are:
Notice that $b$ is negative.

$f(x) = ax^2 + bx + c$

$f(x) = 2x^2 - 8x + 1$

$\begin{aligned}a &= 2\\b&=-8\\c &= 1\end{aligned}$

Step 2

Use the values of $a$, $b$, and $c$ to find the vertex and axis of symmetry.

- Tip: For the formula for the $y$-coordinate of the vertex, substitute the $x$-coordinate of the vertex into the function.
- Use the $x$-coordinate of the vertex for the axis of symmetry.

$\begin{aligned}\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)&=\left(-\frac{-8}{2\cdot 2},1-\frac{(-8)^2}{4 \cdot 2}\right)\\&=\left(\frac{8}{4},1-\frac{64}{8}\right)\\(h,k) &=(2,-7)\end{aligned}$

Solution

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

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