SOLUTION: How do you the parabola y=4x^2-12x+5 into vertex form and intercept form?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: How do you the parabola y=4x^2-12x+5 into vertex form and intercept form?      Log On


   

Question 991566: How do you the parabola y=4x^2-12x+5 into vertex form and intercept form?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Complete the square to put into vertex form,
y=4%28x%5E2-3x%29%2B5
y=4%28x%5E2-3x%2B%283%2F2%29%5E2%29%2B5-4%283%2F2%29%5E2
y=4%28x-3%2F2%29%5E2%2B5-9
y=4%28x-3%2F2%29%5E2-4
Then factor to get to intercept form,
y=%282x-5%29%282x-1%29
.
.
.
.