Question 391299
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)