About Chanda Walker Expertise Any high school level math questions. Please don't just type the math problem without any comments. If you don't tell me what your trouble is, I can't help.
Experience Years of tutoring, BS in mathematics, constant use of mathematics in career.
Finally got around to finishing the course and got my yearend Review. I finished the handout for the most part but got stuck on a few questions and was hoping you could explain them.
When the polynomial x^3-kx^2+17x+6 is divided by x-3, the remainder is 12. What is the value of K?
How would I tackle this? I know that X-3=0. X=3.
Do I plug the 3 in? What do I do with the remainder of 12? I've always had the variables.
Determine the equation of each function in factored form, given the following conditions.
A quartic function tangent to the x-axis at (-1,0) and (2,0) which passes through (3,-16).
Is a quartic function a variable to the power of 4? And when they mention tangent do they mean (-1,0)(-1,0) and (2,0)(2,0)
Hey that wasn't so bad. ;)
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Determine the equation of each function in factored form, given the following conditions.
A quartic function tangent to the x-axis at (-1,0) and (2,0) which passes through (3,-16).
This is a quartic function:
y(x) = ax^4 +bx^3 + cx^2 + dx + e = 0
If it is tangent to the x-axis then we know:
y'(-1) = 0
y'(2) = 0
y'(x) = 4a x^3 + 3b x^2 + 2c x + d
y'(-1) = -4a + 3b - 2c + d = 0 (eq 1)
y'(2) = 32a + 12b + 4c + d = 0 (eq 2)
Since they both equal zero, let's set them equal to each other. I'm going to skip a few steps here . . .
36a + 9b + 6c = 0
We also know
y(-1) = 0
a - b + c - d + e = 0 (eq 3)
y(2) = 0
16a + 8b + 4c + 2d + 3 = 0 (eq 4)
y(3) = -16
81y + 27b + 9c + 3d + e = 16 (eq 5)
That gives us 5 equations and 5 unknowns. From here I think it is simply a linear algebra problem. If you need further help, let me know but I have to attend to other priorities right now.
I don't understand what you are trying to represent with your graph. I'll check back as soon as I can.
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For each pair of functions, find f (g(3)), f (g(x)), g (f(3)), g(f(x))
a. f(x)=5-3x, g(x)=4x+1
For g(3) I just plug in 3 wherever I see X I understand that. But what do I do with f(g(x))?
g(3) = 13
f(g(3)) = f(13) = 5 - 39 = -34
f(g(x)) = 5 - 3(4x + 1) = 5 - 12x - 3 = 2 - 12x
f(3) = -4
g(f(3)) = g(-4) = -15
g(f(x)) = 4(5 - 3x) + 1 = 20 - 12x + 1 = 21 - 12x
Sorry to run out on that second question, I'll be back though!
