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Identify the graph that has a vertex of (-1,1) and a leading coefficient of a=2.

User Yorch
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To determine the vertex form of a parabola has equation:


f(x)=a(x-h)^2+k

where V(h,k) is the vertex of the parabola and 'a' is the leading coefficient.

From the question, we have that, the vertex is (-1, 1)

and the leading coefficient is a = 2

We substitute the vertex and the leading coefficient into the vertex form to

get:


\begin{gathered} f(x)=2(x+1)^2\text{+}1 \\ f(x)=2(x+1)^2+1 \end{gathered}

The graph of this function is shown in the attachment.

Hence the equation of parabola is


f(x)=2(x+1)^2+1

Identify the graph that has a vertex of (-1,1) and a leading coefficient of a=2.-example-1
User KuldipKoradia
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