Given a polynomial function f(x), describe the effects on the y-intercept, regions where the graph is increasing and decreasing, and the end behavior when the following changes are made. make sure to account for even and odd functions. when f(x) becomes f(x) + 2 when f(x) becomes −one half ⋅ f(x)

We are have the function f(x).

Part 1: It is given that the function f(x) is transformed to f(x)+2.

This gives us that the function is translated 2 units upwards.

As, the y-intercept of f(x) is f(0). So, the y-intercept of f(x)+2 is f(0)+2, which is also translated 2 units up.

Translation does not change the behavior of the function.

So, if f(x) is increasing (or deceasing), then f(x)+2 will be increasing(or decreasing) respectively.

Moreover, f(x)+2 will be even or odd if f(x) is even or odd respectively.

Part 2: It is given that the function f(x) is transformed to

This gives us that the function is shrinked vertically by a factor of followed by a reflection across x-axis.

Also, the y-intercept of is .

The graph of f(x) after changing to will flip over the x-axis.

So, the function will increase or decrease when f(x) decrease or increase respectively.

Moreover, will be even if f(x) is even or odd respectively.

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240