Give an example of a rational function (i. e., the quotient of two polynomials) f satisfying the following conditions: • f is not defined at 1. • f(−3) = 0. • f(3) = 9. • lim x→5+ f(x) = −[infinity] and lim x→5− f(x) = [infinity]. explain your reasoning.

6x+18/(-x²+6x-5)

Step-by-step explanation:

• f is not defined at 1

• f(−3) = 0

• f(3) = 9

• lim x→5+ f(x) = −[infinity]

• lim x→5− f(x) = [infinity]

• f is not defined at 1

we need to have a denominator 0 for x = 1

so, 1/(x-1)

• lim x→5+ f(x) = −[infinity]

• lim x→5− f(x) = [infinity]

For the limits, we need to have

1/(-x+5)

this way, lim  x→5+ f(x) = −[infinity] and lim x→5− f(x) = [infinity]

So far we have

1/(x-1)(-x+5)

• f(−3) = 0

the nominator has to be 0 when x = -3

this way, x+3

so, (x+3)/(x-1)(-x+5)

• f(3) = 9

All we need to do is multiply (x+3)/(x-1)(-x+5) by 6, so

6(x+3)/(x-1)(-x+5) = 6x+18/(-x²+6x-5)

dwe

step-by-step explanation:

ewd

shift 4 units up, and 1 unit left.

it will translate f(x) = 3^x to g(x) = 3^(x+1) + 4