Let f(x) = 2x^2 +3x-4 find f(3h) a ) 18h^2 + 9h - 4 b ) 18h^2 + 3h - 4 c ) 2h^2+9h + 14 d) 9h^2 + 9h - 4 do with this one

A

substitute x = 3h into f(x)

f(3h) = 2(3h)² + 3(3h) - 4 = (2 × 9h²) + 9h - 4 = 18h² + 9h - 4

the least value out of the data shown is 5, maybe if you post a picture with it i could a little more.

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! : )

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