Question

Write the equation of the parabola with a vertex at (3, 2) that goes through the point (1,14).

Answer 1

Answer:

$y=2x^2-12x+20$

Explanation:

The vertex form of the equation of a parabola is given as:

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

Given that the vertex (h,k)=(3,2)

We have:

$y=a(x-3)^2+2$

Since it goes through the point (1,14):

x=1, y=14

$\begin{gathered} 14=a(1-3)^2+2 \\ 14=a(-2)^2+2 \\ 14-2=4a \\ 4a=12 \\ a=3 \end{gathered}$

The equation of the parabola is:

$\begin{gathered} y=2(x-3)^2+2 \\ =2(x-3)(x-3)+2 \\ =2(x^2-6x+9)+2 \\ y=2x^2-12x+20 \end{gathered}$

The equation of the parabola is:

$y=2x^2-12x+20$