Question

What is the equation of the parabola that has a vertex at `(-2, -4)` and passes through point `(1,5)`?

Answer 1

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

Explanation:

The equation of a parabola in `color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`
where (h ,k) are the coordinates of the vertex and a is a constant.

`"here " (h,k)=(-2,-4)`

`rArry=a(x-(-2))^2-4`

`rArry=a(x+2)^2-4`

To find a, substitute the point (1 ,5) into the equation. That is x = 1 and y = 5

`rArr5=a(1+2)^2-4`

`rArr9a=9rArra=1`

`"Thus " y=(x+2)^2-4color(red)" is equation in vertex form"`

Expanding the bracket and simplifying gives.

`y=x^2+4x+4-4`

`rArry=x^2+4xcolor(red)" equation in standard form"`