Question

write an equation for a quadratic function with a vertex (0,8) with a parabola that opens down with a dilation of 2

Answer 1

The equation of a parabola is \(ax^2 + bx + c\). \(a\) has been given as \(2\). So we have \(2x^2 + bx + c\). The \(x\)-coordinate of the vertex is \(\dfrac{-b}{2a}\). We know what \(a\) is -- so what is \(b\)?

Answer 2

-b/4

Answer 3

And the x-coordinate of the vertex has been given as \(0\). What is \(b\) equal to?

Answer 4

0?

Answer 5

Yup. So we are left with the following polynomial as \(b\) cancels out:\[2x^2 + c \]When you input \(0\) to that, you get \(8\) because \((0,8)\) is the vertex.

Answer 6

So\[2(0)^2 + c = 8\]Find \(c\).

Answer 7

8

Answer 8

Indeed. So the polynomial is \(2x^2 + 8\).

Answer 9

thanks a bunch

Answer 10

No problem, man!