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Interpreting Quadratic Functions in Vertex Form

Interpreting Quadratic Functions in Vertex Form 1.19 - Solution

We want to write the vertex form of the parabola whose vertex is $(\text{-}4,5)$ and passes through the point $(0,\text{-}10).$ To do so, let's first recall the vertex form of a quadratic function. $\begin{gathered} y={\color{#FF0000}{a}}(x-{\color{#0000FF}{h}})^2+\textcolor{#9400D3}{k} \end{gathered}$ In this form, $({\color{#0000FF}{h}},\textcolor{#9400D3}{k})$ is the vertex of the parabola. Since we are given that the vertex of our function is $({\color{#0000FF}{\text{-}4}},\textcolor{#9400D3}{5}),$ we have that ${\color{#0000FF}{h}}={\color{#0000FF}{\text{-}4}}$ and $\textcolor{#9400D3}{k}=\textcolor{#9400D3}{5}.$ We can use these values to partially write our equation. $\begin{gathered} y={\color{#FF0000}{a}}(x-({\color{#0000FF}{\text{-}4}}))^2+\textcolor{#9400D3}{5} \quad \Leftrightarrow \quad y=a(x+4)^2+5 \end{gathered}$ Finally, to find the value of ${\color{#FF0000}{a}},$ we will use the fact that the function passes through $(0,\text{-}10).$ We can substitute $0$ for $x$ and $\text{-}10$ for $y$ in the above equation and solve for ${\color{#FF0000}{a}}.$
$y=a(x+4)^2+5$
${\color{#009600}{\text{-}10}}=a({\color{#0000FF}{0}}+4)^2+5$
Solve for $a$
$\text{-}10=a\cdot4^2+5$
$\text{-}10=a\cdot16+5$
$\text{-}15=16a$
$\text{-}\dfrac{15}{16}=a$
$a=\text{-}\dfrac{15}{16}$
Now that we know that ${\color{#FF0000}{a}}={\color{#FF0000}{\text{-}\frac{15}{16}}},$ we can complete the equation of the function. $\begin{gathered} y=\text{-}\dfrac{15}{16}(x+4)^2+5 \end{gathered}$