Question

How do you find x intercepts of parabola in vertex form `y=4(x-3)^2-1`?

Answer 1

Develop to standard form:

`f(x) = 4(x^2 - 6x + 9) - 1 = 4x^2 - 24x + 35 = 0.`
To get x-intercepts, solve quadratic equation:
`f(x) = 4x^2 - 24x = 35 = 0 (1)`. There are many methods. I use the transforming Method (Google, Yahoo Search).
Transformed equation: x^2 - 24x + 140. (2) (a.c = 140). Solve (2) by finding 2 numbers p' and q' knowing sum (= -b = 24) and product (c.a = 140).
Compose factor pairs of a.c = 140. Proceed:(2, 70)(4, 35)(5, 28)(10, 14). This last sum is 24 = -b. Then p' = 10 and q' = 14. back to (1) the 2 real roots are: p = p'/a = 10/4 = 5/2; and q = q'/a = 14/4 = 7/2.
x-intercepts or real roots: 5/2 or 7/2.