Question

Write the equation of the quadratic function with roots -1 and 1 and a vertex at (0, -3).

Answer 1

If you have a polynomial with a known set of roots \(r_1, r_2, ... r_n\) you can write the equation in factored form as \[P(x) = a(x-r_1)(x-r_2)...(x-r_n)\]where \(a\) is a constant (usually = 1) that allows you to make the curve pass through an arbitrary point (such as \((0,-3)\) in this case).

Answer 2

Okay can you help me still I don't get that at all

Answer 3

Well, what are your roots?

Answer 4

I don't know what that means

Answer 5

read the problem description again.

Answer 6

-1 and 1

Answer 7

"quadratic function with roots -1 and 1 "

Answer 8

do you understand what a root is?

Answer 9

Yes I do

Answer 10

okay. it's a value of x that makes the polynomial have value 0, right?

Answer 11

yes

Answer 12

and the polynomial will be 0 at each root, and nowhere else, correct?

Answer 13

I think so

Answer 14

\(\bf x=-1\implies x+1=0\implies {\color{blue}{ (x+1)}}=0\\ \quad \\
x=1\implies x-1=0\implies {\color{blue}{ (x-1)}}=0\\ \quad \\

a{\color{blue}{ (x+1)}}{\color{blue}{ (x-1)}}=0\implies a{\color{blue}{ (x+1)}}{\color{blue}{ (x-1)}}=\textit{original polynomial}\)

Answer 15

Omg make is easier

Answer 16

that's correct. so, because a*b = 0 if either a or b = 0 (or both!), we can construct our polynomial from our roots by setting up product terms that will = 0 at each root.

\[(x-1) = 0\]\[(x+1)=0\]those will equal 0 at x = 1, and x = -1

Answer 17

Agreed?

Answer 18

yes

Answer 19

Okay, so that means we can write our polynomial as\[P(x) = a(x+1)(x-1)\]All we have to do now is find the right value of \(a\) so that the polynomial goes through the point \((0,-3)\)

Answer 20

Any idea how we might do that?

Answer 21

No I don't

Answer 22

Well, how about if I write it like this:
\[y = P(x) = a(x+1)(x-1)\]\[(0,-3)\]

Answer 23

we want to find \(a\) such that if we set \(x=0\), we get \(y = -3\)

Answer 24

where do we plug in Y

Answer 25

often these problems just ask for you to write a polynomial with the given roots. There's an infinite number of them, one for each possible value of \(a\). In that case, we just say \(a=1\) and are done. But in this case, we need 1 out of that infinity of choices, so we plug in our known point:
\[y = P(x) = a(x+1)(x-1)\]\[-3 = P(0) = a(0+1)(0-1)\]\[-3 = a(1)(-1)\]

Answer 26

What do you get for \(a\)?

Answer 27

a= 3?

Answer 28

very good! So what is our polynomial?

Answer 29

f(x) = -3x^2 − 3x

Answer 30

?

Answer 31

mmm...no, not quite. try again?

Answer 32

f(x) = 3x^2 + 3

Answer 33

getting closer :-)

Answer 34

f(x) = 3x^2 − 3

Answer 35

Bingo!

Answer 36

can you keep helping me ? I have 9 more question

Answer 37

Here's a plot of that for printing and hanging on the wall after you verify that it crosses the x-axis at x = 1, x = -1, and passes through (0,-3):

Answer 38

can you keep helping me ? I have 9 more question

Answer 39

Yes, but before we go on, I want to beat this horse to death a little bit more :-)

Remember I said there are an infinite number of polynomials that have that same set of roots? Here I've plotted a number of different values of \(a\) (1,2,3,-2). Notice how they all have the same roots, even though the rest of the curve varies?

Answer 40

yes

Answer 41

are you going to help me or not

Answer 42

patience, it's not like I'm getting paid to do this! I'll be back in a few minutes.

Answer 43

promise c;

Answer 44

Okay, the others are finally out of the house, you have my full attention :-)

Answer 45

Find the equation of the quadratic function with zeros 10 and 12 and a vertex at (11, -2). HELP

Answer 46

Okay, this is exactly what we just did. zeros (aka roots) are x = 10, x = 12. What is our equation?

Answer 47

uh I don't know

Answer 48

I think the answer is y = 2x^2 − 44x − 240

Answer 49

Okay, here's the drill again.
To construct a polynomial \(P(x)\) with a given set of roots (or zeros), \(r_1,r_2,...r_n\)
\[P(x) = a(x-r_1)(x-r_2)...(x-r_n)\]If the polynomial must pass through a given point \((x_0,y_0)\), set \[P(x_0) = y_0\]and solve for the necessary value of \(a\), otherwise let \(a = 1\).

Answer 50

I think the answer y = 2x^2 − 44x − 240

Answer 51

Uh, you have an incorrect sign in there, it appears.

Answer 52

y = 2x^2 − 44x + 240

Answer 53

Two tests to see if you've gotten a correct answer (both must be satisfied):

1) evaluating the polynomial at the zeros must give you 0 as a result
2) the polynomial must pass through the target point

\[y = 2x^2-44x+240\]\[2(10)^2-44(10)+240 =200-440+240 = 0\checkmark\]\[2(12)^2-44(12)+240 = 288-528+240 = 0\checkmark\]\[-2 = 2(11)^2-44(11)+240\]\[-2=242-484+240\checkmark\]