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A quadratic function is a polynomial function of degree $2.$ That means that the highest exponent of the independent variable is $2.$ The standard form of a quadratic function is expressed as follows.

$y=ax^2+bx+c$

Here, $a, \, b,$ and $c$ are real numbers and $a \neq 0.$ The simplest quadratic function is $y=x^2,$ and the graph of any quadratic function is a parabola.

The inherent shape of parabolas gives rise to several characteristics that all quadratic functions have in common.

A parabola either opens **upward** or **downward**. This is called its direction.

Because a parabola either opens upward or downward, there is always one point that is the absolute maximum or absolute minimum of the function. This point is called the vertex.

At the vertex, the function changes from increasing to decreasing, or vice versa.

All parabolas are symmetric, meaning there exists a line that divides the graph into two mirror images. For quadratic functions, that line is always parallel to the $y$-axis, and is called the axis of symmetry.

The axis of symmetry always intersects the vertex of the parabola, and is written as a vertical line, where $h$ can be any real number.

$x=h$

Depending on its rule, a parabola can intersect the $x$-axis at $0,$ $1,$ or $2$ points. Since the function's value at an $x$-intercept is always $0,$ these points are called zeros, or sometimes roots.

When a quadratic function is written in standard form, it's possible to use $a,$ $b,$ and $c$ to determine characteristics of its graph. $\begin{aligned} \textbf{direction} &: \text{upward when } a>0, \\ &\phantom{:} \text{downward when } a<0 \\ \mathbf{y} \textbf{-intercept} &: (0,c) \\ \textbf{axis of symmetry} &: x=\text{-}\dfrac{b}{2a} \end{aligned}$

The direction of the graph is determined by the sign of $a.$ To understand why, consider the quadratic function $y=ax^2.$ Since all squares are positive, $x^2$ will always be positive. When $a$ is positive, then $ax^2$ is also positive. Thus, when moving away from the origin in either direction, the graph extends upward. Similarly, when $a$ is negative, $ax^2$ will be negative. Thus, the graph will extend downward for all $x$-values.

The $y$-intercept of a quadratic function is given by $c,$ specifically at $(0,c).$ This is because substituting $x=0$ into standard form yields the following. $\begin{aligned} y&=ax^2+bx+c\\ y&=a \cdot {\color{#0000FF}{0}}^2 + b \cdot {\color{#0000FF}{0}} + c \\ y&= 0+0+c \\ y&=c \end{aligned}$

**Vertex form** is an algebraic format used to express quadratic function rules.

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

In this form, $a$ gives the direction of the parabola. When $a>0,$ the parabola faces upward and when $a<0,$ it faces downward. The vertex of the parabola lies at $(h,k),$ and the axis of symmetry is $x=h.$ Consider the graph of $f(x)=\text{-} \frac{1}{2}(x+4)^2+8.$

From the graph, we can connect the following characterisitcs to the function rule. $\begin{aligned} \textbf{direction} &: \text{downward} &&\rightarrow \quad a<0 \\ \textbf{vertex}&: (\text{-} 4,8) &&\rightarrow \quad h= \text{-} 4, \ k=8\\ \textbf{axis of symmetry}&: x=\text{-} 4 &&\rightarrow \quad h=\text{-} 4 \end{aligned}$

Notice that although the factor in the function rule shows $(x+4)^2,$ $h$ is actually equal to $\text{-} 4.$ This coincides with a horizontal translation of a quadratic function.Quadratic function rules can be expressed in factored form.

$y=a(x-s)(x-t)$

As is the case with standard form and vertex form, $a$ gives the direction of the parabola. When $a>0,$ the parabola faces upward, and when $a<0,$ it faces downward. Additionally, the zeros of the parabola lie at $(s,0) \text{ and } (t,0).$ Because the points of a parabola with the same $y$-coordinate are equidistant from the axis of symmetry, the axis of symmetry lies halfway between the zeros. Consider the graph of $f(x)=(x+4)(x+2).$

From the graph, the following characteristics can be connected to the function rule.

$\begin{aligned} \textbf{direction} &: \text{upward} &&\rightarrow \hspace{-6pt} \quad a>0 \\ \\ \textbf{zeros}&: (\text{-} 4,0) \text{ \& } (\text{-} 2, 0 ) \hspace{-6pt} &&\rightarrow \hspace{-6pt} \quad s=\text{-} 4, \ t=\text{-} 2\\ \\ \textbf{axis of symmetry} &: x=\text{-} 3 &&\rightarrow \hspace{-6pt} \quad x=\frac{\text{-}4 + (\text{-}2)}{2}= \text{-} 3 \end{aligned}$For the following functions, determine the direction, the axis of symmetry, vertex, and zeros. $\begin{aligned} f(x)=(x&+1)(x-5) \quad g(x)=\text{-} 2 (x - 3)^2+2\\[0.3em] &h(x)=0.5x^2+5x+10.5 \end{aligned}$

We'll focus on each function individually, starting with $f.$

$\text{-}2(x-3)^2+2=0$

$\begin{array}{l}x_1=4 \\ x_2=2 \end{array}$

$h(x)=0.5x^2+5x+10.5$

$h({\color{#0000FF}{\text{-}5}})=0.5({\color{#0000FF}{\text{-}5}})^2+5({\color{#0000FF}{\text{-}5}})+10.5$

$h(\text{-}5)=0.5\cdot25-25+10.5$

$h(\text{-}5)=12.5-25+10.5$

$h(\text{-}5)=\text{-}2$

$0.5x^2+5x+10.5$

Solve using the quadratic formula

Use the Quadratic Formula: $a = {\color{#0000FF}{0.5}}, \, b={\color{#009600}{5}}, \, c={\color{#FF0000}{10.5}}$

$x=\dfrac{\text{-}{\color{#009600}{5}}\pm\sqrt{{\color{#009600}{5}}^2-4\cdot{\color{#0000FF}{0.5}}\cdot{\color{#FF0000}{10.5}}}}{2\cdot{\color{#0000FF}{0.5}}}$

$x=\dfrac{\text{-}5\pm\sqrt{25-21}}{1}$

$x=\text{-}5\pm\sqrt{25-21}$

$x=\text{-}5\pm\sqrt{4}$

$x=\text{-}5\pm2$

$\begin{array}{l}x_1=\text{-}3 \\ x_2=\text{-}7 \end{array}$

A function's rate of change gives an indication of how its outputs ($y$) change with respect to its inputs ($x$).

$\text{Rate of Change} = \dfrac{\Delta y}{\Delta x}$

When the rate of change is constant, the function is linear. However, when the function is **not** linear, it's possible to determine an *average rate of change* over an arbitrary interval. This is an increase or decrease between the endpoints of the interval. If the $x$-coordinates of the endpoints are $x_1$ and $x_2,$ and the function is $f,$ the average rate of change is defined as follows.

$\text{Average Rate of Change} = \dfrac{f(x_2)-f(x_1)}{x_2-x_1}$

The function $f(x)$ is quadratic.

Determine the average rate of change over the interval $[\text{-} 3,2].$

To determine the average rate of change, we must know the $x$-values of the endpoints. Since the interval is $[\text{-} 3, 2],$ the $x$-values are $x=\text{-} 3$ and $x=2.$ Let's mark these points on the graph to find their corresponding function values.

The $y$-values are $\text{-}2$ and $\text{-}7.$ Now, we can determine the average rate of change using the formula.$\dfrac{f(x_2)-f(x_1)}{x_2-x_1}$

$\dfrac{f({\color{#009600}{2}})-f({\color{#0000FF}{\text{-}3}})}{{\color{#009600}{2}}-({\color{#0000FF}{\text{-}3}})}$

$\dfrac{{\color{#0000FF}{\text{-}7}}-({\color{#009600}{\text{-}2}})}{2-(\text{-}3)}$

$\dfrac{\text{-}5}{5}$

$\text{-}1$

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