Hello forum.

I have a homework which consists of 40 exercises, but I don't understand it, can you please teach me how to solve these 8 exercises so I could use them as guide to do my H.W, Thanks

1-For the following polynomial funcion f:
f(x)= (x-4)(x+2)^2(x-2)
A- Find the x and y intercepts of F
B Determine whether the graph of f crsses or toches the x axis at each x intercept
C- End behavior. Find the power function that the graph of f resembles for large values of |x|
D- Determine the maximum number of turning points on the graph of f.
E- Use the x intercept(s) to find the intervals on which the graph of f is above and bellow the x axis.
F- Plot the points obtained in parts A and E and use the ramaining information to connect them with a smooth, continuos curve.
2- Find the domain of the given rational functions. Find any vertical, horizontal or oblique asymptotes. Graph the given rational function.
a r(x)=3x^3/x^2-2x-1
b f(x)=)x^2-10x-12/x^2+x-6
c f(x) =1/x-2
d f(x)= x-1/x-2
3-aUse synthetic division to find the quotient q(x) and remainder r(x) when f(x) is divided by g(x). is factor of f? f(x) =x^4-x^2+2x+2 g(x) =x+1
b= use synthetic division to find f(1/2)
c= is (X+1/2) a factor of the function f

4- list all the potential rational zeros of the function f, do not attemtp to find the zeros
f(x) = 12x^8-x^7+6x^4-x^3+x-3
5= use the rational zeros theorem to find all the zeros of each polynomial function. Use the zeros to factor F over the real numbers
f(x) =x^3+2x^2-5x-6
6 form a polynomial f(x) with real coheficients having the giving degree and zeros .Degree 3; zeros:1,-4+i
7- Solve the given equation in the complex number system. x^4+5x^3+2x^2-x+6=0 hint, find p(-3) and p(-2)
8- use the intermediare value theorem to show that each of these polynomials has a zero in the given interval
a f(x) =8x^4-2x^2+5x-1 {0,1}
b f(x) =x^4+8x^3-x^2+2 {-1,0}

thank you.

2. Only had time to do 1A 1B and half of 1C which are just algebra, but I guess it might help if you actually needed those parts.

1

A-
(x-4) (x+2) (x+2) (x-2)

X intercepts = 4, -2, -2, 2.... Whatever makes the stuff in the parenthesis = 0. The two -2 mean that the graph touches the x axis but does not cross. All the other intercepts cross.

To solve for y intercept, set x = 0.

(0-4)(0+2) (0+2) (0-2)=y
-4*2*2*-2=y
y intercept =32
C. End behavior. Remember that a X^2, X^4, X^(even) will always have both ends pointing up. -X^2, -X^4, etc, will have both pointing down
X^1, X^3, X^5, X^7 etc, (x^odd) will have left side pointng down right pnting up.

-x^odd will be left up right down.

Yours will be x^4 something, so both sides should point up.

3. Originally Posted by girl1985
Hello forum.
...
...2- Find the domain of the given rational functions. Find any vertical, horizontal or oblique asymptotes. Graph the given rational function.
a r(x)=3x^3/x^2-2x-1
b f(x)=)x^2-10x-12/x^2+x-6
c f(x) =1/x-2
d f(x)= x-1/x-2
...
Hi,
to a) Calculate the zeros of the denominator to get domain and the vertical asymptotes:
x²-2x-1 = 0 ===> x = 1+√(2) or x = 1 - √(2). Thus d = IR\{1 - √(2), 1 + √(2)}
Do long division to get the oblique asymptote:
(3x³)/(x²-2x-1) = 3x + 6 +(12x-6)/(x²-2x-1). Thus the oblique asymptote has the equation: y = 3x+6
The graph of the function is painted blue, the oblique asymptote in red and the two vertical asymptotes are painted green. Note that the left vertical asymptote nearly covers the graph of the function.

to b) vertical asymptotes at x = -3 and x = 2. Thus d = IR\{-3, 2}
One horizontal asymptote y = 1

to c) vertival asymptote at x = 2, thus d = IR\{2}
One horizontal asymptote y = 0

to d) vertical asymptote at x = 2, thus d = IR\{2}
One horizontal asymptote y = 1

The graphs of c) and d) are missing, but they are easily to sketch

EB

4. Originally Posted by girl1985
...
8- use the intermediare value theorem to show that each of these polynomials has a zero in the given interval
a f(x) =8x^4-2x^2+5x-1 {0,1}
b f(x) =x^4+8x^3-x^2+2 {-1,0}...
Hi,

to a)
calculate f(0) = -1, that means the point B(0, -1) belongs to the graph of the function and is below the x-axis
calculate f(1) = 10, that means the point A(1, 10) belongs to the graph of the function and is above the x-axis.
Thus the graph crosses the x-axis running(?) from B to A. That means the function has a zero: x-value of the interception
Using the theorem: If x increases from 0 to 1 the function has all values between -1 and 10. That means: For a certain x-value there must exist a zero of the function.

to b)
calculate f(-1) = -6, that means the point B(-1, -6) belongs to the graph of the function and is below the x-axis
calculate f(0) = 2, that means the point A(0, 2) belongs to the graph of the function and is above the x-axis.
Thus the graph crosses the x-axis running(?) from B to A. That means the function has a zero: x-value of the interception
Using the theorem: If x increases from -1 to 0 the function has all values between -6 and 2. That means: For a certain x-value there must exist a zero of the function.

EB

5. Originally Posted by girl1985
...
6 form a polynomial f(x) with real coheficients having the giving degree and zeros .Degree 3; zeros:1,-4+i
...
Hi,

if I understand your problem correctly you mean that the function has only real coefficients(?).
If so the complete list of zeros is: 1, -4+i, -4-i

The function f consists of the 3 factors:

f(x) = (x-1)(x+4-i)(x+4+i) Expand all brackets:

f(x) = (x-1)((x+4)² - (-1)) = x³ + 7x² + 9x - 17

EB

6. ## Thank you

Thank you so much for your very helpful assistence, earboth and Terrible Idiot.

please correct the exercise number 3 which I tried to do

a-Use synthetic division to find the quotient q(x) and remainder r(x) when f(x) is divided by g(x). is factor of f? f(x) =x^4-x^2+2x+2 g(x) =x+1
b= use synthetic division to find f(1/2)
c= is (X+1/2) a factor of the function f

a x+1 = -1^4- -1^2+2(-1)+2 = 1-1-2+2 = 0
b -1/2^4 - -1/2^2+2(-1/2)+2 = 1/16-1/4-1+2= 13/16
C (x+1/2) is not a factor of the function f

=======================

I also tried to do exercise number 4

list all the potential rational zeros of the function f, do not attemtp to find the zeros
f(x) = 12x^8-x^7+6x^4-x^3+x-3

I'm not sure if I understand this question correctly, but by using the descartes theorem I assume this equation has at most 3 zeros, since there are 3 changes in the signs. is that correct?

=========
7- Solve the given equation in the complex number system. x^4+5x^3+2x^2-x+6=0 hint, find p(-3) and p(-2)

should I just reeplaze x by the given values?
thank you