SOLUTION: What is the vertex of the parabola given by the equation {{{y=3x^2-48x+115}}}? So far I have completed the square giving me: {{{x^2-16x-192=y+115-576}}}. When I factored the e

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: What is the vertex of the parabola given by the equation {{{y=3x^2-48x+115}}}? So far I have completed the square giving me: {{{x^2-16x-192=y+115-576}}}. When I factored the e      Log On


   

Question 856223: What is the vertex of the parabola given by the equation y=3x%5E2-48x%2B115?

So far I have completed the square giving me: x%5E2-16x-192=y%2B115-576. When I factored the equation I got: %28x-24%29%28x%2B8%29=y-461.
Found 2 solutions by lwsshak3, richwmiller:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
What is the vertex of the parabola given by the equation y=3x%5E2-48x%2B115
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Vertex form of equation: y=A(x-h)^2+k, (h,k)=coordinates of the vertex
y=3x^2-48x+115
complete the square:
y=3(x^2-16x+64)+115-192
y=3(x-8)^2-77
vertex: (8,-77)

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
min{y = 3x^2-48 x+115} = -77 at x = 8
3 x^2-48 x+115
derivative
6x-48
x=8
3*8^2-48*8+115
3*64-48*8+115
192-384+115
y=-77
3x^2-48x=-115
x^2-16x=-115/3
x^2-16x+64=(192-115)/3
3(x-8)^2-77=y
vertex 8,-77
zeroes
x = 8+sqrt(77/3)
x = 8-sqrt(77/3)
parabola
focus | (8, -923/12)=(8, -76.9167)
vertex | (8, -77)
semi-axis length | 1/12=0.0833333
focal parameter | 1/6=0.166667
eccentricity | 1
directrix | y = -925/12