SOLUTION: How to find the x and y intercepts and the vertex of f(x)=2x^2+12x-3

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Question 751928: How to find the x and y intercepts and the vertex of f(x)=2x^2+12x-3
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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How to find the x and y intercepts and the vertex of f(x)=2x^2+12x-3
Write it
y = 2x^2 + 12x - 3
The x intercept is when y = 0
2x^2 + 12x - 3 = 0
Use the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
Where: a=2; b=12; c=-3
You should get x intercepts of
x ~ .24
and
x ~ -6.24
:
The y intercept occurs when x = 0
y = 2(0^2) + 12(0) - 3
y = -3 is the y intercept
:
Vertex occurs on the axis of symmetry: x = -b/(2a)
x = %28-12%29%2F%282%2A2%29
x = -3 is the axis symmetry
Find y, when x = -3
y = 2(-3^2) + 12(-3) - 3
y = 18 - 36 - 3
y = -21
:
The vertex: -3,-21