algebra 2 problem

Question

f(x) = -(x-3)2(x+1)3(x-7)
Given the following polynomial, find
a. The zeros and multiplicity of each
b. Where the graph crosses or touches the axis
c. Number of turning points
d. End behavior
In part b, if the real zero has even multiplicity (i.e., the exponent associated with it is even), the graph touches the x-axis at that point, but does not cross; if it has odd multiplicity (i.e., the exponent associated with it is odd), the graph crosses the x-axis at that point. So the intent is to look at the multiplicity of each zero, and determine whether that multiplicity is an even or odd number.
Since "1" is an odd number, we would say that the graph would cross the x-axis at x = 2.
Since "2" and "4" are even numbers, we would say that the graph would touch the x-axis at x =-3 and 1/2.
In part c, the maximum number of turning points is found by imagining what would happen if we multiplied the function out into a polynomial:
f(x) = 5(x - 2)(x + 3)²(x - 1/2)^4
= (5x - 10)(x² + 6x + 9)(x - 1/2)(x - 1/2)(x - 1/2)(x - 1/2)
= (5x^3 + 20x^2 – 15x – 90)(x^2 – x + ¼)(x^2 – x + ¼)
= (5x^3 + 20x^2 – 15x – 90)(x^4 – 2x^3 – 3/2x^2 – 1/2x + 1/16)
= 5x^7+ 10x^6 – 62.5x^5 – 92.5x^4 – 212.5x^3 – 212.2x^2 + 44.1x – 5.6
Specifically, what is the largest exponent n of the variable x that we could possibly get from that product? The maximum number of turning points will be n-1.
In this case, the largest exponent we can get is “7”; so the maximum number of turning points is 7 – 1 = 6.
Part d is very similar. Assuming that we multiplied the function out as in part c above, what would be the leading term of that product? That is, what would be the term in that product that contained the largest exponent of x? That leading term is called the power function; and it dictates the end behavior of the function for large positive and negative values of x, since the values of the other terms would be negligible by comparison.

In this case, the power function would be x7.
Let’s look at another example.
Let's say we were given
f(x) = 3(x – 2)^2(x + 2)^2
For part a, we set
3(x – 2)^2(x + 2)^2 = 0
and find the solutions are x = 2, -2. Since there is an exponent of 2 applied to each quantity, these roots both have multiplicity 2, which is an even number; so the graph touches the x-axis at x = 2, -2, but does not cross that axis at these points.
For part c, we have
f(x) = 3(x – 2)^2(x + 2)^2
= 3(x – 2)(x + 2)(x – 2)(x + 2)
= 3(x2 – 4)(x^2 – 4)
= 3x4 – 24x^2 + 48
The largest exponent of x that we can get in this product will be in the term 3x4; so there will be 4-1 = 3 turning points.
For part d, we repeat that the term with the largest exponent of x that we can get in this product will be 3x4; so that is the power function.

Is there any way I can get some assistance in solving this?

Answer 1

zeros are 3, -1 and 7

also = the x intercepts, where the graph crosses the x axis, at 3, -1 and 7

multiplicity of 2, 3 and 1 for 3, -1 and 7 respectively

= the powers of each factor

crosses at -1 and 7, but just touches, is tangent at 3

turning points at 0, 3 and about 6

it's a 6 degree polynomial when multiplied out, with potentially 1 less turning point 6-1 =5 maximum possible

use a graphing calculator