Question

Find the quadratic function that y=f(x) that has the vertex (0,0) and whose graph passes through the point(-2,-8). Write the function in standard form

Answer 1

Given

The quadratic function that y=f(x) that has the vertex (0,0) and whose graph passes through the point(-2,-8)

Solution

Recall

`y=a(x-h)^2+k`

`\begin{gathered} Vertex\text{ =\lparen h,k\rparen} \\ \\ h=0 \\ k=0 \\ \\ y=a(x-0)^2+0 \\ y=ax^2 \end{gathered}`

Given

Point (-2, -8)

`\begin{gathered} x=-2 \\ y=-8 \\ -8=a(-2)^2 \\ -8=a4 \\ -8=4a \\ divide\text{ both sides by 4} \\ (4a)/(4)=-(8)/(4) \\ \\ a=-2 \end{gathered}`

Now

`\begin{gathered} y=a(x-h)^2+k \\ a=-2 \\ h=0 \\ k=0 \\ y=-2(x-0)^2+0 \\ y=-2(x)^2+0 \\ y=-2x^2 \end{gathered}`

The standard form

`y=ax^2+bx+c`

Now

`\begin{gathered} y=-2x^2+0x+0 \\ which\text{ is } \\ y=-2x^2 \end{gathered}`

Checking with graph

The final answer

`y=-2x^2`