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
[tex]y=a(x-h)^2+k[/tex][tex]\begin{gathered} Vertex\text{ =\lparen h,k\rparen} \\ \\ h=0 \\ k=0 \\ \\ y=a(x-0)^2+0 \\ y=ax^2 \end{gathered}[/tex]Given
Point (-2, -8)
[tex]\begin{gathered} x=-2 \\ y=-8 \\ -8=a(-2)^2 \\ -8=a4 \\ -8=4a \\ divide\text{ both sides by 4} \\ \frac{4a}{4}=-\frac{8}{4} \\ \\ a=-2 \end{gathered}[/tex]Now
[tex]\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}[/tex]The standard form
[tex]y=ax^2+bx+c[/tex]Now
[tex]\begin{gathered} y=-2x^2+0x+0 \\ which\text{ is } \\ y=-2x^2 \end{gathered}[/tex]Checking with graph
The final answer
[tex]y=-2x^2[/tex]
