Question

the graph of f(x)=x³ with three reference points is shown below. (0.5 point) 4. which of the following represents points from f(x) to a point on g(x)=-3(x + 2)³-2? select three that apply. a. (-3,1) b. (0,-2) c. (-2,-2) d. (-1,-5) e. (1,-2) learning goal from lesson 5.2 i can find the key features of and then graph the polynomial functions (zeros via factorization and end behavior) how i did (circle one) i got it! im still learning it. 5. which graph is sketched to the right? (0.5 point) a. g(x)=x(x - 2)(x + 1) b. g(x)=x(x + 2)(x - 1) c. g(x)=x(x - 2)(x - 1)

Explanation:

4.

Step1: Recall transformation rules

If \(y = f(x)\) and \(y = a(x - h)^{3}+k\), the transformation from \(f(x)=x^{3}\) to \(g(x)=-3(x + 2)^{3}-27\) involves a horizontal shift of \(h=- 2\) units to the left, a vertical stretch by a factor of \(|a| = 3\), a reflection over the \(x\) - axis (\(a=-3\)), and a vertical shift of \(k = - 27\) units down.
For a point \((x_0,y_0)\) on \(f(x)=x^{3}\), the corresponding point \((x,y)\) on \(g(x)\) is given by \(x=x_0 - 2\) and \(y=-3y_0-27\).
We can also check by substituting \(x\) - values into \(g(x)=-3(x + 2)^{3}-27\).

Step2: Check option A

For \(x=-3\) in \(g(x)=-3(x + 2)^{3}-27\), we have \(g(-3)=-3(-3 + 2)^{3}-27=-3(-1)^{3}-27=3 - 27=-24
eq1\).

Step3: Check option B

For \(x = 0\) in \(g(x)=-3(x + 2)^{3}-27\), we have \(g(0)=-3(0 + 2)^{3}-27=-3\times8-27=-24 - 27=-51
eq - 2\).

Step4: Check option C

For \(x=-2\) in \(g(x)=-3(x + 2)^{3}-27\), we have \(g(-2)=-3(-2 + 2)^{3}-27=-27
eq - 2\).

Step5: Check option D

For \(x=-1\) in \(g(x)=-3(x + 2)^{3}-27\), we have \(g(-1)=-3(-1 + 2)^{3}-27=-3\times1-27=-3 - 27=-30
eq - 5\).

Step6: Check option E

For \(x = 1\) in \(g(x)=-3(x + 2)^{3}-27\), we have \(g(1)=-3(1 + 2)^{3}-27=-3\times27-27=-81-27=-108
eq - 2\).

5.

Step1: Find the zeros of the polynomial functions

For \(A. g(x)=x(x - 2)(x + 1)\), the zeros are \(x = 0\), \(x=2\) and \(x=-1\).
For \(B. g(x)=x(x + 2)(x - 1)\), the zeros are \(x = 0\), \(x=-2\) and \(x = 1\).
For \(C. g(x)=x(x - 2)(x - 1)\), the zeros are \(x = 0\), \(x=2\) and \(x = 1\).
The graph has zeros at \(x=-2\), \(x = 0\) and \(x = 1\). So the function is \(g(x)=x(x + 2)(x - 1)\)

Answer:

1. None of the options are correct.
2. B. \(g(x)=x(x + 2)(x - 1)\)