Question

Find the equation for a parabola that has a vertex at (0, 4) and passes through the point (–1, 6).

Answer 1

Answer:

$\large\boxed{y=2x^2+4}$

Step-by-step explanation:

The vertex form of a parabola:

$y=a(x-h)^2+k$

(h, k) - vertex

We have the vertex at (0, 4) → h = 0 and k = 4.

Substitute:

$y=a(x-0)^2+4=ax^2+4$

The point (-1, 6) is on the parabola. Put the coordinates of the point to the equation:

$6=a(-1)^2+4$            subtract 4 from both sides

$2=a\to a=2$

Finally:

$y=2x^2+4$