The function h(x) = 12x8 + 49 is an even function. Which transformation of h(x) would result in a function that is neither even nor odd? reflection over the x-axis vertical stretch by a factor of 7 translation 8 units to the right horizontal compression by a factor ofOne-half

 translation 8 units to the right

Step-by-step explanation:

An even function is a function such that:

h(x) = h(-x)

and the function is odd if:

h(-x) = -h(x)

Now, let's talk about transformations:

A) Reflection over the x-axis:

When we have a point (x,y) and we do a reflection over the x-axis, the reflected point will be:

(x, -y).

Then for the case of a function:

y = h(x).

then the reflection will be:

g(x) = - y = -h(x).

And if h(x) is even, -h(x) is also even, so this is not the correct option.

B) Vertical stretch by a factor of 7.

This is written as:

g(x) = 7*h(x).

then:

g(x) = 7*h(x)

g(-x) = 7*h(-x) = 7*h(x) = g(x)

the transformation is even.

C) Translation of 8 units to the right.

We can write this as:

g(x) = h(x - 8).

Then:

g(-x) = h(-x -8) = h(x + 8)

Then g(-x) is not equal to g(x)

and also g(-x) ≠ -g(x)

So in this case the transformation is neither odd or even.

C)  horizontal compression by a factor of One-half.

This transformation is written as:

g(x) = h( (1/2)*x)

And, similar as the case of the vertical compression, in this case the transformation is also even.

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

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry