College algebra 1

Question

f(x)=-3x^2-18x-28. Where does the maximum value occur? What is the functions maximum value

Answer 1

The function f(x) is a quadratic function and its leading coefficient is -3 (the number in the front of x^2) is less than 0. So it has a maximum value. The maximum value appears at its vertex which is given by (-b/2a, f(-b/2a)); to be precise, the maximum appears when x=-b/2a and the maximum value is f(-b/2a). In this case, a=-3 (the coefficient of x^2) and b=-18 (the coefficient of x). Therefore, -b/2a = -(-18)/2(-3)=18/(-6)=-3 and f(-b/2a)=f(-3)=-3(-3)^2-18(-3)-28=-1. Now we can conclude that the maximum of f(x) occurs at x=-3 and the maximum value itself is -1.

The other way to approach this question is to use the method of completing squares:

f(x) = (-3x^2-18x) -28 (Group the first two terms involving x)

=-3(x^2+6x) - 28 (Factor out the common factor -3)

=-3(x^2+6x+(6/2)^2 - (6/2)^2) -28 (Add and subtract (6/2)^2 at the same time)

= -3(x^2+6x+9-9)-28 (Simplify (6/2)^2)

=-3[(x+3)^2-9]-28 (x^2+6x+9 becomes (x+3)^2 which is a square)

=-3(x+3)^2+27-28 (Distribute -3)

=-3(x+3)^2 - 1 (Simplify)

Note that -3(x+3)^2<=0. So f(x) obtains the max when -3(x+3)^2=0 or x=-3. The maximum value is -1.

Answer 2

Given: y = f(x) = - 3x^2 - 18x - 28

Find: Maximum and where it occurs.

NOTE: This is a parabola

With negative leading coefficient.

So, this parabola open downward

and has a maximum at the vertex.

Vertex occurs on the axis of symmetry

x = -b/(2a)

x = [ -(-18) ] / [ 2(-3) ]

x = [ 18 ] / [ -6 ]

x = -3

Axis of symmetry is at: x = -3

Vertex occurs at ( -3, f(-3) )

f(x) = - 3x^2 - 18x - 28

f(-3) = - 3(-3)^2 - 18(-3) - 28

f(-3) = -3 (9) + 54 - 28

f(-3) = -27 + 54 - 28

f(-3) = -1

Vertex occurs at (-3, -1)

The maximum value of the function is -1

The maximum value occurs at x = -3