SOLUTION: Y=1/2x^2+5x+1 find the vertex, x intercepts and axis of symmetry

Question 1027110: Y=1/2x^2+5x+1 find the vertex, x intercepts and axis of symmetry
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The vertex has an x-value of -b/2a
Here, that is -5/1=-5 for x and y is +12.5-25+1=-11.5
(-5,13.5)
x-inercepts--set y=0
1/2 x^2+5x+1=0
multiply by 2 to clear fractions
x^2+10x+2=0
x=(1/2)(-10 +/- sqrt (100-8)); sqrt 92=2 sqrt(23)
x=-5+/- sqrt (23); numerically, it is -0.204 and -9.80
Axis of symmetry is around the vertex, and that is at -5.

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