The graph of f(x) = |x| is translated 6 units to the right and 2 units up to form a new function. Which statement about the range of both functions is true?

Option C.The range changes from {y | y > 0} to {y | y > 2}.

Step-by-step explanation:

The complete question is

The graph of f(x) = |x| is translated 6 units to the right and 2 units up to form a new function. Which statement about the range of both functions is true?

A.The range is the same for both functions: {y | y is a real number}.\

B.The range is the same for both functions: {y | y > 0}.

C.The range changes from {y | y > 0} to {y | y > 2}.

D.The range changes from {y | y > 0} to {y | y > 6}.

we have

The vertex of this function is the point (0,0)

The range of f(x) is the interval [0,∞)

If f(x) is is translated 6 units to the right and 2 units up to form a new function g(x)

then

The rule of the translation of f(x) to g(x) is

(x,y) ----> (x+6,y+2)

The new vertex is

(0,0) ---> (0+6,0+2)

(0,0) ---> (6,2)

so

The new function equation is

The range of g(x) is the interval [2,∞)

therefore

The statement that is true is

The range changes from {y | y > 0} to {y | y > 2}.

answer: howdy

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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