SOLUTION: find -intercepts -vertex -max or min -range g(x)=x^2-6x+5

Question 355143: find -intercepts -vertex -max or min -range
g(x)=x^2-6x+5
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
Hi,
g(x)=x^2-6x+5
.
Factoring to find the intercepts
g(x)=x^2-6x+5 = (x - 5)(x - 1)
.
Intercepts (5,0) and (1, 0)
.
*Note: Complete the square to put into the vertex form of an equation of a parabola y = a(x - h)^2 + k where (h,k) is the vertex
.
(completing the square by adding and subtracting 9)
g(x) = x^2-6x+5 = (x^2 - 6x + 9) -9 + 5= (x-3)^2 -4
.
vertex = (3, -4) this is the minimum
(a = 1, parabola opens upward)
.
Range is (,)
.

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