Expand menu menu_open Minimize Start chapters Home History history History expand_more
{{ item.displayTitle }}
navigate_next
No history yet!
Progress & Statistics equalizer Progress expand_more
Student
navigate_next
Teacher
navigate_next
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
No results
{{ searchError }}
search
menu_open
{{ courseTrack.displayTitle }}
{{ statistics.percent }}% Sign in to view progress
{{ printedBook.courseTrack.name }} {{ printedBook.name }}
search Use offline Tools apps
Login account_circle menu_open

Interpreting Quadratic Functions in Vertex Form

Interpreting Quadratic Functions in Vertex Form 1.19 - Solution

arrow_back Return to Interpreting Quadratic Functions in Vertex Form
We want to write the vertex form of the parabola whose vertex is (-4,5)(\text{-}4,5) and passes through the point (0,-10).(0,\text{-}10). To do so, let's first recall the vertex form of a quadratic function. y=a(xh)2+k\begin{gathered} y={\color{#FF0000}{a}}(x-{\color{#0000FF}{h}})^2+\textcolor{#9400D3}{k} \end{gathered} In this form, (h,k)({\color{#0000FF}{h}},\textcolor{#9400D3}{k}) is the vertex of the parabola. Since we are given that the vertex of our function is (-4,5),({\color{#0000FF}{\text{-}4}},\textcolor{#9400D3}{5}), we have that h=-4{\color{#0000FF}{h}}={\color{#0000FF}{\text{-}4}} and k=5.\textcolor{#9400D3}{k}=\textcolor{#9400D3}{5}. We can use these values to partially write our equation. y=a(x(-4))2+5y=a(x+4)2+5\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 a,{\color{#FF0000}{a}}, we will use the fact that the function passes through (0,-10).(0,\text{-}10). We can substitute 00 for xx and -10\text{-}10 for yy in the above equation and solve for a.{\color{#FF0000}{a}}.
y=a(x+4)2+5y=a(x+4)^2+5
-10=a(0+4)2+5{\color{#009600}{\text{-}10}}=a({\color{#0000FF}{0}}+4)^2+5
Solve for aa
-10=a42+5\text{-}10=a\cdot4^2+5
-10=a16+5\text{-}10=a\cdot16+5
-15=16a\text{-}15=16a
-1516=a\text{-}\dfrac{15}{16}=a
a=-1516a=\text{-}\dfrac{15}{16}
Now that we know that a=-1516,{\color{#FF0000}{a}}={\color{#FF0000}{\text{-}\frac{15}{16}}}, we can complete the equation of the function. y=-1516(x+4)2+5\begin{gathered} y=\text{-}\dfrac{15}{16}(x+4)^2+5 \end{gathered}