SOLUTION: 1/2x^2 + x -1=0 ax^2 + bx + c=0

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Question 182219: 1/2x^2 + x -1=0



ax^2 + bx + c=0
Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!

Given system%28%281%2F2%29x%5E2+%2B+x+-1=0%29
f%28x%29=ax%5E2%2Bbx%2Bc, where a, b, and c are real numbers, with a%3C%3E0
Remember:system%28a=1%2F2%2Cb=1%2Cc=-1%29
To find x-coordinate of the vertex--->x=-b%2F%282a%29
x=-1%2F%282%2A%281%2F2%29%29=-1%2F%28cross%282%29%2A1%2Fcross%282%29%29
highlight%28x=-1%29
To find y-Intercept:
y=%281%2F2%29%28-1%29%5E2%2B%28-1%29-1=%281%2F2%29-2=%281-4%29%2F2=-3%2F2
highlight%28y=-3%2F2%29
Vertex---->(-1,-3/2)
The Axis of symmettry is x=-1.
Solving for X-Intercepts:
By Quadratic formula:

x=%28-1%2B-sqrt%281%2B2%29%29%2F1=%28-1%2B-sqrt%283%29%29%2F1
x=%28-1%2B1.732%29%2F1=highlight%280.732%29
x=%28-1-1.732%29%2F1=highlight%28-2.732%29
To find Y-Intercept:
f%28x%29=0
y=%281%2F2%29%280%29%5E2%2B0-1=highlight%28-1%29
We'll mark all highlighted points in the graph:
---->
Thank you,
Jojo