Question

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

Answer 1

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`