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Interpreting Quadratic Functions in Factored Form

Interpreting Quadratic Functions in Factored Form 1.11 - Solution

To draw the graph of the given function, we will follow four steps.

1. Identify and plot the $x\text{-}$intercepts.
2. Find and graph the axis of symmetry.
3. Find and plot the vertex.
4. Draw the parabola through the vertex and the points where the $x\text{-}$intercepts occur.

Let's go through these steps one at a time.

Identify and Plot the $x\text{-}$intercepts

Recall the factored form of a quadratic function. $\begin{gathered} y=a(x-p)(x-q) \end{gathered}$ In this form, where $a$ $\neq$ $0,$ the $x\text{-}$intercepts are $p$ and $q.$ Let's consider the factored form of our function. $\begin{gathered} f(x)=\text{-}3(x+1)(x+7) \\ \Updownarrow \\ f(x)={\color{#FF0000}{\text{-}3}}(x-({\color{#0000FF}{\text{-} 1}}))(x-({\color{#009600}{\text{-}7}})) \end{gathered}$ We can see that ${\color{#FF0000}{a}}={\color{#FF0000}{\text{-}3}},$ ${\color{#0000FF}{p}}={\color{#0000FF}{\text{-}1}},$ and ${\color{#009600}{q}}={\color{#009600}{\text{-}7}}.$ Therefore, the $x\text{-}$intercepts occur at $({\color{#0000FF}{\text{-}1}},0)$ and $({\color{#009600}{\text{-}7}},0).$

Find and Graph the Axis of Symmetry

The axis of symmetry is halfway between $(p,0)$ and $(q,0).$ Since we know that $p=\text{-}1$ and $q=\text{-}7,$ the axis of symmetry of our parabola is halfway between $(\text{-}1,0)$ and $(\text{-}7,0).$ $\begin{gathered} x=\dfrac{p+q}{2}\quad \Rightarrow \quad x=\dfrac{{\color{#0000FF}{\text{-}1}}+({\color{#009600}{\text{-}7}})}{2}=\dfrac{\text{-}8}{2}=\text{-}4 \end{gathered}$ We found that the axis of symmetry is the vertical line $x=\text{-}4.$

Find and Plot the Vertex

Since the vertex lies on the axis of symmetry, its $x\text{-}$coordinate is $\text{-}4.$ To find the $y\text{-}$coordinate, we will substitute $\text{-}4$ for $x$ in the given equation.
$f(x)=\text{-}3(x+1)(x+7)$
$f({\color{#0000FF}{\text{-}4}})=\text{-}3({\color{#0000FF}{\text{-}4}}+1)({\color{#0000FF}{\text{-}4}}+7)$
$f(\text{-}4)=\text{-}3(\text{-}3)(3)$
$f(\text{-}4)=9(3)$
$f(\text{-}4)=27$
The $y\text{-}$coordinate of the vertex is $27.$ Therefore, the vertex is the point $(\text{-}4,27).$

Draw the Parabola

Finally, we will draw the parabola through the vertex and the $x\text{-}$intercepts.