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Interpreting Quadratic Functions in Vertex Form

Interpreting Quadratic Functions in Vertex Form 1.18 - Solution

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We are given four quadratic functions to compare and find the one which doesn't belong. y1=8(x+4)2y2=(x2)2+4y3=2x2+x+4y4=3(x+1)2+1\begin{aligned} &y_1 = 8(x+4)^2 & &y_2 = (x-2)^2+4 \\ &y_3 = 2x^2+x+4 & &y_4 = 3(x+1)^2+1 \end{aligned} We can see some functions are written in vertex form. Therefore, it will be useful to recall how to identify the vertex from this notation.

Vertex Formy=a(xh)2+k\begin{gathered} \underline{\textbf{Vertex Form}}\\ y = \textcolor{#FF00FF}{a}(x-{\color{#0000FF}{h}})^2 + {\color{#009600}{k}} \end{gathered} In this format, the vertex is located at (h,k).({\color{#0000FF}{h}},{\color{#009600}{k}}).

Let's identify the vertex of each of the given functions.

Function Vertex
y1=8(x+4)2y_1 = 8(x+4)^2
\Updownarrow
y1=8(x(-4))2+0y_1 = \textcolor{#FF00FF}{8}(x-({\color{#0000FF}{\text{-} 4}}))^2 + {\color{#009600}{0}}
(-4,0)({\color{#0000FF}{\text{-} 4}},{\color{#009600}{0}})
y2=(x2)2+4y_2 = (x-2)^2+4
\Updownarrow
y2=1(x2)2+4y_2 = \textcolor{#FF00FF}{1}(x-{\color{#0000FF}{2}})^2 + {\color{#009600}{4}}
(2,4)({\color{#0000FF}{2}},{\color{#009600}{4}})
y3=2x2+x+4y_3 = 2x^2+x+4 Not possible with this information
y4=3(x+1)2+1y_4 = 3(x+1)^2+1
\Updownarrow
y4=3(x(-1))2+1y_4 = \textcolor{#FF00FF}{3}(x-({\color{#0000FF}{\text{-} 1}}))^2 + {\color{#009600}{1}}
(-1,1)({\color{#0000FF}{\text{-} 1}},{\color{#009600}{1}})

As we can see, function y3=2x2+x+4y_3 = 2x^2+x+4 is the only one that is not written in vertex form.