Question

The parábola y = ax^2+k has a vertex (0,-7) and passes through the point (2,4) find the equation Y=

Answer 1

Given:

The vertex of the parabola, (h, k) = (0, -7)

The parabola passes through the point (x, y) = (2,4).

To find the parabola equation:

The general form is,

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

Substitute h=0, k=-7, x=2, and y=4. We get,

`\begin{gathered} 4=a(2-0)^2-7 \\ 4=4a-7 \\ 4a=11 \\ a=(11)/(4) \end{gathered}`

Substitute the values of a, h, and k in the general form.

We get,

`\begin{gathered} y=(11)/(4)(x-0)^2_{}-7 \\ y=(11)/(4)x^2_{}-7 \end{gathered}`

Hence, the parabola equation is,

`y=(11)/(4)x^2_{}-7`