SOLUTION: What formula do you use to put this in vertex form where you can graph it? What are the "zeros" of this equation? and how do you find the "zeros". f(x)=-3(x-2)^2+5

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: What formula do you use to put this in vertex form where you can graph it? What are the "zeros" of this equation? and how do you find the "zeros". f(x)=-3(x-2)^2+5      Log On


   

Question 123842: What formula do you use to put this in vertex form where you can graph it? What are the "zeros" of this equation? and how do you find the "zeros".

f(x)=-3(x-2)^2+5
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
A zero means to let f(x) = 0 and solve for x:

f(x)=-3(x-2)^2 + 5 =0
Subtract 5 from each side:
-3%28x-2%29%5E2++=-+5

Divide by -3:
+%28x-2%29%5E2=5%2F3
Take the squsre root of each side:
x-2=0%2B-sqrt%285%2F3%29

Add 2 to each side:
x=2%2B-sqrt%285%2F3%29+

There are other ways to write this, by rationalizing the denominator:
x=2%2B-+sqrt%28%285%2F3%29%2A%283%2F3%29%29

x=2%2B-+sqrt%2815%2F9%29
x=2%2B-+%28sqrt%2815%29%29%2F3

You may need to express this with a common denominator:
+x=%286%2B-sqrt%2815%29%29%2F3

R^2