Ihave somewhat of an idea on how to do this but i'm not 100% sure

Check out the attached below to see the final graph.

We have 3 pieces that make up the piecewise function.

First up is f(x) = x when x <= -2, meaning that we only graph y = x when x is -2 or smaller. This part is shown in red in the graph below. There is a closed circle at the endpoint due to the "or equal to" as part of the inequality. The "Less than" means we draw the left portion or everything to the left of x = -2.

The next piece is f(x) = x^2-2 and it's only graphed if -2 < x < 0. So if x is between -2 and 0, not including either endpoint, then we graph out y = x^2-2, otherwise we don't graph it. The blue curve is the second piece. The endpoints of the blue curve are open circles. We don't include these points as part of the function.

Finally, we have the purple piece on the graph which is y = (1/2)x-2 when x >= 0. If x is 0 or larger, then we graph y = (1/2)x-2. This piece connects with the endpoint of the blue piece to cover up or fill in the second open hole.

So the final result is what you see below in the attached image.

so then what are you asking?

step-by-step explanation:

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source