SOLUTION: what is the equation for finding the vertex to y=(1/4)(x^2-2x+5)

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Question 117654: what is the equation for finding the vertex to y=(1/4)(x^2-2x+5)
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

y=%281%2F4%29%28x%5E2-2x%2B5%29

y=%281%2F4%29x%5E2-+%281%2F4%292x%2B+%281%2F4%295
y=%281%2F4%29x%5E2-+%281%2F2%29x%2B+%285%2F4%29…….=>…a=1%2F4,b=1%2F2,and c=5%2F4

To find the value of “x” at the vertex, use this formula: x+=+-%28b%2F2a%29
x+=+-%28%281%2F2%29%2F%282%281%2F4%29%29%29

x+=+-%28%281%2F2%29%2F%282%2F4%29%29

x+=+-%28%281%2F2%29%2F%281%2F2%29%29

x+=+-%282%2F2%29

x+=+-+1


to find y, substitute your answer for x+ into the original equation to find the y, value of the vertex

y=%281%2F4%29%28x%5E2-2x%2B5%29…………….. x+=+-+1

y=%281%2F4%29%281%5E2-2%2A1+%2B5%29

y=%281%2F4%29%281+-2+%2B5%29

y=%281%2F4%29%284%29

y=+1
So, the “point” of the vertex is: (-1, 1). This is the "POINT" where the quadratic equation will "change" direction on the graph.