SOLUTION: Find the x-intercepts and the vertex of the parabola. f(x) = 4x^2

SOLUTION: Find the x-intercepts and the vertex of the parabola. f(x) = 4x^2 - 56x + 160

Question 82952: Find the x-intercepts and the vertex of the parabola.
f(x) = 4x^2 - 56x + 160
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Find the x-intercepts and the vertex of the parabola.
:
f(x) = 4x^2 - 56x + 160
y = f(x)
y = 4x^2 - 56x + 160
:
you can simplify it by dividing equation by 4, you have:
x^2 - 14x + 40 = 0
:
Factors easily to:
(x - 10)(x - 4) = 0
x Intercepts:
x = +10
and
x = +4
:
Find the axis of symmetry: x = -b/(2a), in this equation a=1; b=-14
x = -(-14)/(2*1)
x = +14/+2
x = + 7 is the axis of symmetry
:
Use x = 7 to find the vertex.
You have to find the vertex using the original equation, (not the simplified eq)
:
Substitute 7 for x in y = 4x^2 - 56x + 160
y = 4(7^2) - 56(7) + 160
y = 4(49) - 392 + 160
y = 196 - 392 + 160
y = -36
:
The vertex x/y coordinates are +7, -36
:
Did this make sense to you? Any questions?
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