Question

help please =)

Answer 1

help with what?

Answer 2

@whpalmer4?

Answer 3

y = (x+3)^2(x-1)^2
To find y-intercept, set x = 0:
y = (0+3)^2 * (0-1)^2 = 9 * 1 = 9
To find x-intercept set y = 0 and solve for x:
0 = (x+3)^2(x-1)^2
What values of x will make the right hand side zero?

Answer 4

idk =(

Answer 5

The right hand side will be zero IF either of the factor is zero.
That is either (x+3)^2 = 0 or (x-1)^2 = 0
(x+3)^2 = 0 when x = -3
(x-1)^2 = 0 when x = 1
Therefore, the x-intercepts are: -3 and 1.

Answer 6

y = (x+3)^2(x-1)^2
Can you tell what the degree of this polynomial is?

Answer 7

2?

Answer 8

It is 4 because x is squared in (x+3)^2 and it is multiplied by another term where x is once again squared in (x-1)^2. So the highest degree term is x^4.
For large values of |x|, the function will behave very much like the highest degree term.
Therefore for b), y = x^4 is the power function that f resembles for large values of |x|

Answer 9

A polynomial of degree "n" will have a maximum of (n-1) turning points.
Since f(x) = (x+3)^2(x-1)^2 is a polynomial of degree 4, it will have a maximum of (4-1) which is 3 turning points. Answer to c) 3

Answer 10

d)
f(x) = (x+3)^2(x-1)^2 = (x+3)(x+3)(x-1)(x-1)
At x = -3, f(x) = 0. Since x+3 occurs twice, the root x = -3 has a multiplicity of 2.
At x = 1, f(x) = 0. Since x-1 occurs twice, the root x = 1 also has a multiplicity of 2.

Since the degree of this polynomial is 4 which is even, and each root has a multiplicity of 2, the function will just bounce off the x-axis at x = -3 and at x = 1. "bounce off" meaning it will not cross the x-axis but just touch the x -axis and turn back as shown in the graph below: