Question

Write a quadratic function that fits the given set of points.

Answer 1

(0, 1), (2, -1), and (3, -8)

Answer 2

\[f(x)=ax^2+bx+c\] and substitute your points and find a,b,c

Answer 3

R: the general form of the quadratic function is \[y=ax ^{2}+bx+c.\]
You are given 3 points, each of which must satisfy that equation. You don't know the values of a, b or c; the goal of the problem is to determine those values. Since there are 3 points, you will end up with three equations in 3 unknowns; in other words, this problem is solvable and will produce a unique solution.

Example: With point (2,-1), the equation becomes -1=a(2)^2+b(2)+c.
Substitute the coordinates of each of the other points into the general form of the quad. eqn.; that will give you 2 more equations in terms of a, b and c.

You end up with a system of linear equations.

Methods of solution include: addition/subtraction, substitution, matrices, determinants and more.

Think about this problem. Are you able to proceed on your own now, or have you questions you'd like to ask first?

Answer 4

Be certain to simplify each of the 3 equations. That -1=a(2)^2+b(2)+c becomes:

4a + 2b + c = -1, and both of the other two equations should be put into that form.

Answer 5

y = ax^2 + bx + c

\[0 = a(1)^2 + b(1) + c \]
\[-1 = a(2)^2 + b(2) + c\]
\[-8 = a(3)^2 + b(3) + c\]


Right?

Answer 6

While you're absolutely on the right track, I won't be checking every detail. I suggest you simplify each of the 3 equations as shown, and then attempt to solve the resulting system of linear equations. Which of the methods I listed is most familiar and comfortable to you?

Answer 7

I am still having some trouble understanding how to get it into the other format.

Answer 8

OK, then I'll take another look at what you've typed.
Rather than do all that typing again, I'll just comment on each of your equations.
The first one simplifies to 0=1a+1b+c, or (better), 1a + 1b + 1c = 0.
The second one simplifies to 4a + 2b + c + -1 (because (2)^2 = 4.

Answer 9

The third equation simplifies to
9a +3b + c = -1 because (3)^2=9.

Answer 10

oh okay, and then once those are all sorted out, what do I do from there?

Answer 11

So now, Ricardo, you have a system of linear equations:
1a + 1b +1c = 0
4a + 2b +1c = -1
9a + 3b +1c = -8

We'll have to solve that system for {a, b, c}.

I've been at the computer for about three hours straight, so would like to get off now. But before I do that, have you any comments or questions about what we've done so far and what we need to do next?

Answer 12

Nope! I think I got it all! thank you!

Answer 13

To answer your question: Have you solved systems of linear equations before?
If so, using which method?
Again, the commonest are "addition/subtraction," substitution, determinants and matrices. Personally, I'd choose matrices, but that's because I have a TI-83 calculator on hand and it will solve systems like this very quickly.

Answer 14

OK. If you're satisfied now, fine; we'll stop. You don't have to answer my last questions (above).

Answer 15

Okay, I found a couple of online calculators that can help! So from here I think I have got this pretty under control! Tahnk you for all your help today @ mathmale

Answer 16

OK: One more thing: The solution to this system is a=-3, b=8, c=-5.

Answer 17

Bye!

Answer 18

Haha! thank you!

Answer 19

One last comment: I've tried checking my result, and have found that there's possibly a mistake somewhere. So don't fret if you don't get the same result (solution) as I. Rather, check your own result by substituting the x-coordinates of the given 3 points into the equation y=ax^2+bx+c and determining whether the calculated y value is the same as the given y-coordinate.