final pratice exam help please no calculator

Question

Answer 1

24).

Profit equation is P(x) = - 5x² + 10x + 20.

To find the maximum profit, first find P '(x).

P(x) = - 5x² + 10x + 20

P '(x) = - 10x + 10

Equate P '(x) to 0.

P '(x) = - 10x + 10 = 0

10x = 10

x = 1.

At x = 1, P(1) = - 5(1)² + 10(1) + 20

= - 5 + 10 + 20

= 5 + 20

= 25.

This is the maximum profit.

Solution : Option d is the correct choice.

Comments

what if i want to find the minium or is there no such thing and never will ask that? I am only in pre calc

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The profit equation is P(x) = - 5x² + 10x + 20.

Minimum and maximum values of quadratic functions

Consider the profit of the function with vertex .

1. If , f  has a minimum at and the minimum value is .

2. If , f  has a maximum at and the maximum value is .

Compare the above function with .

a = - 5, b = 10, and c = 20.

Here, a = - 5 < 0, so, the function doesn' t have the minimum value.

Answer 2

25).

The function is f( x ) = - x².

The graph of f(x) horizontally 3 units to the left means : g(x) = - (x + 3)².

and vertically translating the result (- (x + 3)² ) 2 units up means : g(x) = - (x + 3)² + 2.

Therefore, the function g(x) = -(x + 3)² + 2.

g(- 2) = -(- 2 + 3)² + 2

= - (1)² + 2

= - 1 + 2

= 1.

Solution : Option d is the correct choice.

Comments

so if its to the left its +, if its to the right its - when its horizontally. When its vertically + means up, and - means down. Am I right?

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Yes, you are right.