Mathematics, 14.07.2019 12:30 glyptictriton9575

Which of the following is the function for the graph below and shows the end behavior of the function as x ⟶ +∞? f(x) = (4x^2 + 7x - 2)(2x + 3); f(x) ⟶ -∞ f(x) = (4x^2 - 7x - 2)(2x - 3); f(x) ⟶ +∞ f(x) = (4x^2 - 7x + 2)(2x - 3); f(x) ⟶ +∞ f(x) = (4x^2 - 7x - 2)(2x + 3); f(x) ⟶ -∞

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The function for the graph is f(x) = (4x^2 - 7x - 2)(2x - 3) and shows the end behavior of the function as x ⟶ +∞, f(x) ⟶ +∞

Option B

Given :

The graph of a function . Lets analyze the end behavior of the given graph

When x approaches infinity on the right the graph goes up . It means the graph goes to + infinity

As x-> +∞, y -> +∞

So we look at the options B and C

Now, from the graph we can see that y intercept is (0,6)

Lets check options B and C that given y intercept (0,6)

Replace x with 0 and find out f(0)

$f(x) = (4x^2 - 7x - 2)(2x - 3)\\x=0\\f(0) = (4(0)^2 - 7(0)- 2)(2(0)- 3)=6\\y intercept is (0,6)$

the function for the graph is f(x) = (4x^2 - 7x - 2)(2x - 3) and shows the end behavior of the function as x ⟶ +∞, f(x) ⟶ +∞

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Answer from: shadow6728g

The simplest solution:

y - intercept of the function on the graph is (0; 6).

Only function f(x) = (4x^2 - 7x - 2)(2x - 3) has that y-intercept

f(0) = (4 · 0² - 7 · 0 - 2)(2 · 0 - 3) = (-2)(-3) = 6

Other functions have y - intercept equal (0; -6).

f(0) = (0 ... - 2)(0 + 3) = (-2)(3) = -6 or f(0) = (0 ... + 2)(0 - 3) = (2)(-3) = -6

Answer from: Quest

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

Answer from: Quest

sure i'm up for it

step-by-step explanation:

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