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Question 391299: write the function f(x)=4x^2-48x-141 in vertex form, and identify its vertex
Answer by haileytucki(390)   (Show Source):
You can put this solution on YOUR website! f(x)=4x^(2)-48x-141
Replace f(x) with y to find the properties of the parabola.
y=4x^(2)-48x-141
To create a trinomial square on the left-hand side of the equation, add a value to both sides of the equation that is equal to the square of half the coefficient of x. In this problem, add (-6)^(2) to both sides of the equation.
y=4(x^(2)-12x+36)+4(-(141)/(4))-(4)(0+36)
Remove the 0 from the polynomial; adding or subtracting 0 does not change the value of the expression.
y=4(x^(2)-12x+36)+4(-(141)/(4))-(4)(36)
Factor the perfect trinomial square into (x-6)^(2).
y=4((x-6)^(2))+4(-(141)/(4))-(4)(36)
Factor the perfect trinomial square into (x-6)^(2).
y=4(x-6)^(2)+4(-(141)/(4))-(4)(36)
Multiply 4 by each term inside the parentheses.
y=4(x-6)^(2)-141-(4)(36)
Multiply 4 by 36 to get 144.
y=4(x-6)^(2)-141-(144)
Multiply -1 by the 144 inside the parentheses.
y=4(x-6)^(2)-141-144
Subtract 144 from -141 to get -285.
y=4(x-6)^(2)-285
This is the form of a paraboloa. Use this form to determine the values used to find vertex and x-y intercepts.
y=a(x-h)^(2)+k
Use the standard form to determine the vertex and x-y intercepts.
a=4_k=-285_h=6
The vertex of a parabola is (h,k).
Vertex: (6,-285)
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