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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

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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

Find the quadratic function that y=f(x) that has the vertex (0,0) and whose graph-example-1
User Paul Whittaker
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