SOLUTION: what is the intercepts, axis of symmetry and vertex of y=3x^2+12x+16

Question 251606: what is the intercepts, axis of symmetry and vertex of y=3x^2+12x+16
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
what is the intercepts, axis of symmetry and vertex of y=3x^2+12x+16
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y-intercept: Let x = 0, then y = 16
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x-intercepts:
Solve 3x^2+12x+16 = 0
x = [-12 +- sqrt(144 - 4*3*16)]/6
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x = [-12 +- sqrt(-48)]/6
Since the discriminant is negative there are no x-intercepts.
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Vertex occurs where x = -b/2a = -12/(2*3) = -2
f(-2) = 3(4)+12(-2)+16 = 28-24 = 4
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Vertex (-2,4)
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Axis of symmetry is x=-b/2a = -2
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Cheers,
Stan H.
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