What is the range of the function f(x) = 3x2 + 6x – 8? {y|y ≥ –1} {y|y ≤ –1} {y|y ≥ –11} {y|y ≤ –11}

The range of the function f(x)=3x²+6x-8 is:

{y|y ≥ –11}

Step-by-step explanation:

The function f(x)=3x²+6x-8 could also be written as:

f(x)=3x²+6x+3-11

f(x)=3(x²+2x+1)-11

f(x)=3(x+1)²-11

as we know that the value of 3(x+1)²≥0

Then, the value of 3(x+1)²-11≥-11

i.e. f(x)≥-11

Hence, the range of the function f(x)=3x²+6x-8 is:

{y|y ≥ –11}

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

1. sometimes

2. never true

3. sometimes

step-by-step explanation:

these are pretty self explanatory if you know what supp right and com angles are