jmc

Question

Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,1). Express your answer in the form "ax^2+bx+c".

Accepted Answer

Since the axis of symmetry is vertical and the vertex is

(2, 4)

, the parabola may also be written as

y = a (x - 2)^{2} + 4

for some value of

a

. Plugging the point

(1, 1)

into this expression gives

1 = a (1 - 2)^{2} + 4 = a + 4.

This tells us

a = - 3

.

Our equation is

y = - 3 (x - 2)^{2} + 4.

Putting it

y = a x^{2} + b x + c

form requires expanding the square, so we get

y = - 3 (x^{2} - 4 x + 4) + 4 = - 3 x^{2} + 12 x - 8 .