Transformations

Question

A quadratic function has zeros −3 and 5 and contains the point (4, 14).

1. Find the equation of the quadratic function in standard form.
2. Complete the square to determine the vertex of the function.
3. Find the equation of a second quadratic function in the same family as the function in part a) that contains the point (1, −48).

Answer 1

(x+3)(x-5) = x^2 -2x -15

y = 14, when x = 4

f(x) = 14 = a(x^2 -2x -

14 = a (-7)

a = -14/7 - -2

a = -2

f(x) = -2(x^2 -2x -15)

f(x) - -2(x+3)(x-5)

put the equation in vertex form to find the vertex

y = -2x^2 +4x +30

y = -2(x^2 -2x ) +30

complete the square, add 1, if you add 1 inside the parentheses, balance it by subtracting 1(-2) or adding 2

y = -2(x^2 -2x + 1) + 30 +2

y = -2(x-1)^2 + 32

vertex is (1, 32)