Asked by Johnny
identify the vertex of the parabola y=4x2-40x+96
Answers
Answered by
Reiny
several ways to do this.
quickest way:
the x of the vertex is -b/(2a)
= -(-40)/8 = 5
sub back into equation
y = 4(25) -40(5) + 96 = -4
vertex (5,-4)
calculus way:
dy/dx = 8x - 40 = 0
x = 5
proceed as above
completing the square:
y = 4x^2 - 40x + 96
= 4(x^2 - 10x + <b>25 - 25</b> + 96
= 4((x-5)^2 - 25) + 96
= 4(x-5)^2 - 100 + 96
=4(x-5)^2 - 4
quickest way:
the x of the vertex is -b/(2a)
= -(-40)/8 = 5
sub back into equation
y = 4(25) -40(5) + 96 = -4
vertex (5,-4)
calculus way:
dy/dx = 8x - 40 = 0
x = 5
proceed as above
completing the square:
y = 4x^2 - 40x + 96
= 4(x^2 - 10x + <b>25 - 25</b> + 96
= 4((x-5)^2 - 25) + 96
= 4(x-5)^2 - 100 + 96
=4(x-5)^2 - 4
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