Question

the graph of a function is a parabola that has a minimum at (-1,-2) and goes through the point (0,1) what is the equation of the function in the standard form

Answer 1

Given:

Vertex at (-1,-2)

Goes through point (0,1)

The vertex form of a quadratic equation is defined as

`\begin{gathered} y=a(x-h)^2+k \\ \text{where} \\ (h,k)\text{ is the vertex of the equation} \end{gathered}`

Substitute (h,k) with the vertex (-1,-2)

`\begin{gathered} y=a(x-h)^2+k \\ y=a(x-(-1))^2+(-2) \\ y=a(x+1)^2-2 \end{gathered}`

Now we solve for a, by substituting (x,y) with (0,1) to the vertex form of the previous equation.

`\begin{gathered} y=a(x+1)^2-2 \\ 1=a(0+1)^2-2_{} \\ 1=a(1)^2-2 \\ 1=a(1)-2 \\ 1=a-2 \\ 1+2=a \\ a=3 \end{gathered}`

Putting it together with the vertex form we have

`y=3(x+1)^2-2`

Expand the equation, and change it into standard form

`\begin{gathered} y=3(x+1)^2-2 \\ y=3(x^2+2x+1)-2 \\ y=3x^2+6x+3-2 \end{gathered}`

Simplify further, and the equation of the parabola in standard form is

`y=3x^2+6x+1`