Question

Help? Find a cubic function with the given zeros.

-7, 5, -3

Answer 1

The basic form of a cubic function is y=a(x-b)(x-c)(x-b) where a is a constant, and b,c,d are the zeros.

Answer 2

Thank you! Is that last parentheses supposed to be (x-d)?

Answer 3

yes, sorry about that.

Answer 4

you get to pick any value you like for a — 1 would be the obvious one here. if you had to make the function pass through some arbitrary point, a value other than 1 might be needed.

Answer 5

Oh, that makes sense. Thanks you guys! God bless!

Answer 6

the tricky part is multiplying the sucker out! also, don't forget that it is \(x-r\) where \(r\) is the root or zero (interchangeable terminology), so that will be subtracting a negative number for two of your three product terms.

Answer 7

if your zeros are -1,-2, and 3, your product would be \[(x-(-1))(x-(-2))(x-3) = (x+1)(x+2)(x-3)\]

Answer 8

You multiply each part of one, times each part of the other, right? I got: \[x^3+5x^2-29x-105\]

Answer 9

Yep, that's exactly right!

Answer 10

Here's a related topic: if we had to factor that polynomial you just got, we could make guesses as to what the zeros would be by factoring the last term (-105).

Answer 11

notice that -7*-3*5 = 105?

Answer 12

the last term is always the product of the last terms of all the factors (*)

Answer 13

Awesome! Thank you for that equation, I had never seen that before. :)

Answer 14

Oh my gosh. No, I didn't see that. You're a genius! Thank you!

Answer 15

(*) that's assuming the leading term has a 1 for the coefficient. It's a bit more complicated than that if it's something other than 1. this is called the rational root theorem.

Answer 16

Haha. Awesome. Do you know how to find a vertex?

Answer 17

no matter how hairy the polynomial is, if it has some rational zeros, you can get an idea of what they might be by factoring the coefficients of the first and last terms. Unfortunately, you'll end up with more possibilities than actual zeros in most cases, but at least it is a starting point!

Answer 18

vertex of a parabola?

Answer 19

I brought the RRT up only because you'd just done that multiplication that serves as such a fine example! As you said, most people don't notice that property until it is pointed out to them...

Answer 20

Now when your teacher finally gets around to the RRT, you can lean back and tell your friends that you've known about it for almost forever :-)

Answer 21

Like, of this: f(x) = 2x^2 - 8x + 9

Answer 22

Yay! Lol.

Answer 23

Yeah, that's a parabola. \[f(x) = ax^2 + bx + c\] is the so-called "standard form" for a parabola.

Answer 24

there are a number of ways you can tackle this. One is to rearrange the equation into to "vertex form" where you can just read out the coordinates of the vertex. Vertex form is
\[f(x) = a(x-h)^2 + k\] where the vertex is at \((h,k)\)

Answer 25

Another way is to use the property of a parabola in standard form that the vertex is located at \[x = -\frac{b}{2a}\]

Answer 26

We find the value of \(x\) and plug it back into the equation and find the corresponding value of \(y\).

Answer 27

Oh my gosh. I literally had the vertex equation written down on the top of my paper. I'm an idiot. So, for my equation, the vertex is 2, right?

Answer 28

If you know calculus, you can just look at the equation and say "the x value of the vertex is the value of \(x\) where \(4x-8 = 0\)" :-)

Answer 29

The x-value is 2, yes. Now you plug x=2 into \[y=2x^2-8x+9=2(2)^2-8(2)+9 = \]

Answer 30

That gives you one.. What is that?

Answer 31

That's the y-coordinate of the vertex. So the full specification of the vertex of that parabola is (2,1), and you can see that in the attached graph.

Answer 32

Got it. Thanks. :D

Answer 33

so the calculus thing I mentioned: imagine you had a marble rolling around inside that parabola. where is it going to finally stop?

Answer 34

I have no idea. Sorry, I'm doing a test.

Answer 35

down at the bottom, right? that's where the curve is perfectly level, if only for an infinitesimally small distance. calculus gives us a tool called the derivative, which lets us write a function describing the slope of a wiggly curve at any point along the curve. the vertex of a parabola is a point where the slope = 0. calculus tells us that function for the slope of that parabola is 4x-8, so when the value of 4x-8=0, x is the location of the vertex.

Answer 36

good luck with the test.

Answer 37

Hey, do you know how many real or imaginary zeros this would have? f(x) = x^4 - 15x^2 - 16

Answer 38

you always have as many zeros as the highest power in the polynomial.

Answer 39

How do you know how many will be imaginary? Sorry, should have rephrased that question.

Answer 40

Well, do you know about Descartes rule of signs?

Answer 41

There are no changes f sign in this one, so are there no imaginary zeros?

Answer 42

How would I find the domain of 16/13-x?

Answer 43

Write out your polynomial in descending order (highest power of x first)

count the number of times the coefficients change sign when going from left to right
that gives you the maximum number of positive real roots. it may be reduced by a multiple of 2, so long as it doesn't go negative.

now rewrite your polynomial in the same order, except change the sign on any term with an odd-number as the power

repeat the counting procedure with this new poly. that give you the maximum number of negative real roots. again, this number may be reduced by a multiple of 2, so long as it doesn't go negative.

finally, the imaginary roots make up whatever is left over from the maximum exponent. if you have a polynomial with x^5 as the highest power, for example, the positive real roots + negative real roots + imaginary roots = 5.

Answer 44

x^4-15x^2-16 changes sign from +1 to -15 — that's 1 positive real root
x^4-15x^2-16 changes sign from +1 to -15 — that's 1 negative real root
(just happens that we didn't have any x^3 or x terms, so f(x) = f(-x))

Answer 45

\[\frac{16}{13-x}\]can you think of any values that \(x\) cannot be?

Answer 46

Nevermind, it was AR#'s except 13, right?

Answer 47

Hey, how would I do the horizontal asymptote of this? Well, I'll just put the first part, because the exponent powers are the same. 7x^2/2x^2

Answer 48

yes, at x=13, that becomes a division by 0

Answer 49

Would it just be 7/2 @whpalmer4 ?