Question

Y=x^2+8x+1 in vertex form

Answer 1

Given:

`y=x^2+8x+1`

The vertex form of a quadratic equation is given by;

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

Using a graph plotter to get the vertex of the parabola,

The vertex is at (-4,-15) and the curve passes through the point (0,1)

Hence,

`\begin{gathered} h=-4,k=-15 \\ x=0,y=1 \end{gathered}`

Substituting these points into the equation for the vertex form of a parabola;

`\begin{gathered} y=a(x-h)^2+k \\ y=a(x-(-4))^2+(-15) \\ y=a(x+4)^2-15 \\ \\ Substitute\text{ the point to get the value of a} \\ 1=a(0+4)^2-15 \\ 1=16a-15 \\ 1+15=16a \\ 16=16a \\ (16)/(16)=a \\ a=1 \\ \\ Hence,\text{ the equation in vertex form is;} \\ y=1(x+4)^2-15 \\ y=(x+4)^2-15 \end{gathered}`

Therefore, the equation in vertex form is;

`y=(x+4)^2-15`