SOLUTION: Finding the vertex and x and y intercepts of: G(t)=1/2(t^2-4t)? I know I can turn this into vertex form like so: G(t)=1/2(t^2-4t+4)-4 G(t)=1/2(t-2)^2-4 so then the vertex

Algebra ->  Functions -> SOLUTION: Finding the vertex and x and y intercepts of: G(t)=1/2(t^2-4t)? I know I can turn this into vertex form like so: G(t)=1/2(t^2-4t+4)-4 G(t)=1/2(t-2)^2-4 so then the vertex       Log On


   

Question 948282: Finding the vertex and x and y intercepts of: G(t)=1/2(t^2-4t)?
I know I can turn this into vertex form like so:
G(t)=1/2(t^2-4t+4)-4
G(t)=1/2(t-2)^2-4
so then the vertex would be
(2,-4) but graphing it online it states that the vertex is:(2,-2) what am I missing here? what makes it a -2 instead of a -4?
Please also explain how to find the x and y intercepts thank you
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
When you added the 4 in the parentheses, you were really adding %281%2F2%294=2.
Well you subtracted 4 but it should have been 2.
So,
G%28t%29=%281%2F2%29%28t%5E2-4t%2B4%29-2
G%28t%29=%281%2F2%29%28t-2%29%5E2-2
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X-intercepts: Set y=0 solve for x.
0=%281%2F2%29%28t%5E2-4t%29
0=%281%2F2%29%28t%28t-4%29%29
So then t=0 and t=4 are the x-intercepts (or the t-intercepts).
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Y-intercept : Set t=0 solve for y.
y=0
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