PLEASE help me! Algebra.. I'm lost!?

Question

My teacher did not explain this well, and now Im stuck! PLEASE help me understand!!

The first five problems concern the following two functions:

f(x) = x2 + x -30
g(x) = x3 - 3x2 + 4x - 2



Problems 1-5
Algebraically, find the following and compare with the graph.
1. The exact value(s) of x when f(x) = 0

2. The exact value(s) of x when g(x) = 0

For #1 and #2, include any imaginary roots.

3. Use a graphing calculator to find the approximate (x,y) coordinates, to three decimal places, of any intersections of the two functions.

4. Algebraically find the exact (x,y) coordinates of any maximums or minimums of either function.

5. In your own words and in complete sentences, tell what the intersection of the two functions represents.

Answer 1

Mention : I use * for multiply



this is a grade 2 equation

you should know the basics.

take it easier for each care take a, b and c out



x*x+x-30=0

in the first case

a=1

b=1

c=-30



and you also have to calculate D(delta - sorry, I have no greek letters) = b*b - 4*c*a

in this case it is 1- (4*-30)

=1+120=121

fortunately the square root is an exact number =11



the equation is x1,2(both values of x)= (-b + or - square root of D)/4a= (-1 + or - 11)/4

so x1 = -3

x2=5/2=2.5



2. we have to transform it into a grade 2 equation(i didn't do 3 grade equations at school yet)

ANd I have no idea :D



I haven't made graphics for 2 grade equations so I can't help you anymore, sorry, I'm only an 8th grader



5. If I were to guess I would say 2 or 4 points.

Answer 2

Okay.



1. You could plug this equation straight into the quadratic formula, which is x = (-b plus or minus sqrt(b^2 - 4ac)) / 2a. I'm going to give you the faster way.



You can factor x^2 + x - 30 = (x + 6)(x - 5).



You set f(x) equal to zero, so 0 = (x + 6)(x - 5)



That means either 0 = x + 6 or 0 = x - 5. For the first equation x = -6, and for the second equation x = 5.



ANSWER 1: x = 5 or -6



2. This has a similar idea. Using long division, you can factor x^3 - 3x^2 + 4x - 2 = (x - 1)(x^2 - 2x + 2).



Then you set g(x) equal to 0, so 0 = (x - 1)(x^2 - 2x + 2).



That means 0 = x - 1 or 0 = x^2 - 2x + 2. The first equation is easy because x = 1. The second equation cannot be factored, and you have to use the quadratic formula. If you're familiar with the formula, you can just skip to the answer.



The formula is x = (-b plus or minus sqrt(b^2 - 4ac)) / 2a.



You plug in the numbers, so x = (-1 * -2 plus or minus sqrt((-2)^2 - 4 * 1 * 2) / (2 * 1 ).



You clean up the multiplication to get x = (2 plus or minus sqrt(4 - 8)) / 2.



You set 4 - 8 as -4, so x = (2 plus or minus sqrt(-4)) / 2.



The square root of -4 is an imaginary number 2i, so x = (2 plus or minus 2i) / 2.



You divide everything by 2, so x = 1 plus or minus i.



ANSWER: x = 1, 1 + i, or 1 - i.



3. Your teacher says to use your graphing calculator, so feel free to do so. If it's a TI model, there's a button that says Y=. Enter your equations there. Then hit 2ND and CALC, and press enter a couple times.



ANSWER: x = -1.934 and y = -28.19.



4. I'll deal with each function on its own.



f(x) is a parabola. Parabolas only have one minimum and no maximums. If it's negative, reverse what I just said.



The equation for finding the minimum is x = -b / 2a. The reason why comes from calculus.



So, plug in your numbers to get x = -1 * 1 / (2 * -30).



Then, you clean up your multiplication, so x = -1 / -60.



You eliminate the negative signs to get x = 1/60.



You put that into your original equation get y, so y = (1/60)^2 + 1/60 - 30.



You carry out the exponent to get y = 1/3600 + 1/60 - 30.



Then, put everything into a common denominator, so y = 1/3600 + 60/3600 - 108000/3600.



You gather everything in the numerator, so y = -107939/3600. You can't simplify it anymore, so that's the answer to f(x).



On the other hand, g(x) is a cubic function, which has the form g(x) = ax^3 + bx^2 + cx + d. The coordinates of your maximums and minimums are given by the function 0 = 3ax^2 + 2bx + c. The reason why comes from calculus again.



So, first you plug in the numbers to get 0 = 3x^2 - 6x + 4.



Then, you apply the quadratic formula, so x = (-1 * -6 plus or minus sqrt((-6)^2 - 4 * 3 * 4)) / (2 * 3).



Clean up the multiplication to get x = (6 plus or minus sqrt(36 - 48)) / 6.



36 - 48 is -12, and the square root of -12 is 2i * sqrt(3), so x = (6 plus or minus 2i * sqrt(3)) / 6.



Reduce the fractions to get x = 1 plus or minus 2i * sqrt(3). Since these are imaginary answers, there are no minimums or maximums for g(x).



ANSWER: f(x) has one minimum at x = 1/60 and y = -107939/3600.

g(x) has no minimums or maximums.



5. Intersections of two functions mean several things. Check your textbook for what your teacher wants.

Answer 3

i think you can enter that on the calculator(ti-84) idk... your acually kinda cheating...u should try opening a book or paying attention in class.....