Find the quadratic function f(x) = ax² + bx + c for which ƒ( −...

Question

Find the quadratic function f(x) = ax² + bx + c for which ƒ( − 2) = −4, ƒ(1) = 2, and f(2) = 0.

Accepted Answer

Welcome back. I am so glad you're here. We are asked to identify this quadratic function which is A X squared plus B. X plus C. And that equals f of X. So what does that mean to identify it? Well, we need to solve for a. B and C. And how do we do that? Well, we're actually given three pairs of X and F of X coordinates. We're told that F of two. Well, that's our X equals negative three. That's our F of X. So we can write an equation for that. Our first equation here that A stays. But then that X we said that was too, so eight times two squared plus the B stays and that X is still two plus C equals the F of X, which was the corresponding negative three. There's one equation, we've got another set of coordinates. We are given that F of one, that's our X equals two. That's our F of X. So we can set up a second equation A times r X is one squared plus B, times X is one plus C equals in the f of X is two. And then our third equation corresponds with this pair here negative one is our X for f of negative one equals 00 is our F of X. So this third equation would be eight times negative one squared plus B times negative one plus C equals r F of X zero. So we've got a system of equations. We can simplify this a little bit. So this would be the first equation two squared is four, so it would be four. A plus two, B plus C equals negative three. The second equation would be one squared is one, so a plus B times one is B plus C equals two. And the third equation would be negative one squared is a positive one, so that's still A and then B times negative one is minus B plus C equals zero. And look at our system of equations. Now we recall from previous lessons that we can take the coefficients and constant terms from our system of equations. We can turn that into a matrix. And so our corresponding matrix, the first equation would be our first row, coefficients are 4-1 in the constant term is -3. And then the second row, the second equation would be 1112. And the third equation, the third row would be one negative 110. And we've got a matrix and we recall that we can use Gaussian or Gauss Jordan elimination to solve for r, A B and C. So what we can do is we want to get this matrix looking like it has diagonal ones for the first three columns. And then I'm going to use Gauss Jordan elimination, which means we'll have zeros in the rest of the spaces for the first three columns. And then our fourth column will give us our a B and C solutions. So if you're solid on converting these matrix these matrices, this matrix great. We'll see at the end checking our answer. If you need some guidance for converting that, then I'm going to show you every step. It's going to take a little while but we can do it. All right. So the first step to get this blue matrix looking like the red matrix, is that we want this four to become a one. So we're going to start off by replacing row one and in order to do that, there are many different ways. But I like to put off fractions as long as I possibly can. So I'm going to take the second row and I'm going to multiply it by negative three. So we'll have a negative three times row two and add that to row one. And really the only time you can get fancy with that is the first time because we don't have any of our zeros and ones in place yet. So because we're replacing row one, I'm going to copy down Rose two and three. So that's 11121 negative 110. And now doing the work to replace row one, negative three times one is negative three plus four is the one that we want negative three times one is negative three plus two is negative one, negative three times one is negative three plus one is negative two and negative three times two is negative six plus negative three is negative nine. Okay, now we need to have zeros in the rest of column one. So let's work on row two. We'll replace row two. And to do that to get a zero there, we're going to work with the row that has the one in the desired position which was row one. And in this case we'll multiply row one times a negative one And add that to row two because we're replacing road to We need to copy over Rose one and three. I'll do that one negative one, negative two negative nine one negative 110. Now doing the work to replace row two negative one times one is negative one plus one is the zero that we want negative one times negative one is positive one plus one is two negative one. Times negative two is a positive two plus one is three negative one times negative nine is positive nine plus two is 11. All right. Now finishing off column one. We want a zero in this third row. We'll replace row three in order to do that. We'll work with the row that has the one in the desired position which is still row one. And again we will multiply. Row one times a negative one and then add that to row three. Because we're replacing row three. We need to copy over rows one and two. So I'll do that one, negative one, negative two negative 9023, 11. And now doing the work to replace row three negative one times one is negative one plus one is the zero that we want negative one times negative one is positive one plus negative one is another zero negative one times negative two is positive two plus one is a positive three and negative one times negative nine is a positive nine plus zero is nine. Okay. The first column is done. Now we need to get a one in the desired position in column two which is where this two is. So we're going to be replacing road to in order to do that Now is when we have to stick with just working on that road to get that into a one. We're going to multiply it by its reciprocal. So we're going to take 1/2 times row two. And because we're replacing road to I need to copy over Rose one and three. So that will be one negative one negative two negative 90039. Now replacing road to zero times one half is 02 times one half is the one that we want. Three times one half is three halves. There's those fractions and 11 times one half is halves. Okay, now we want zeros in the rest of column two. We've already got one on the bottom. Yeah, so now let's just work on row one replacing row one in order to get a zero into that position. We work with the road that has the one in the desired position which is row two. And in this case we just need to add row two to row one. Because we're replacing row one, we need to copy over rows two and 013 halves, 11 halves and 0039. Now replacing row one and adding road to to it. Zero plus one is 11 plus negative one is the zero that we want. Ye three halves plus negative two is negative one half and 11 halves plus negative nine is negative seven halves. Okay, column two is done. Now we want a one in the desired position of column three. That's where this three is will be replacing row three in order to do that. We're going to multiply row three times the reciprocal of that three. So times one third one third times row three. Because we're replacing row three, we can copy over rows one and 210 negative one half negative seven halves, 013 halves and 11 halves. And now doing the work to replace row 30 times one third is zero, both times three times one. Third is the one that we want and nine times one third is three. Look at that, we're almost done. We need to get zeros in the rest of the positions in column three. So let's work on row one, replacing this negative one half in order to get that to be a zero. We're going to work with the row that has the one in the desired position. So row three let's multiply row three times one half And add that to row one. Because we're replacing row one, we need to copy over Rose two and three. So 013 halves, 11 halves, 0013. And now replacing row 10 times one half is zero plus one is 10 times one half. Zero plus zero is 01 times one half is one half plus negative one half is the zero that we want. Ye and three times one half is three halves plus negative seven halves is negative two. All right. Now this last one that we need to do. We need to get this three halves to be a zero. So we're replacing row two in order to get a zero there we work with the row that has the one in the desired position, which is row three. Let's multiply row three times negative three halves And then add that to row two. Because we're replacing road to we can copy over Rose one and three. So we've got 100, negative two and 0013. Now doing the work to replace row two, zero times negative three halves. Zero plus zero is 00 times negative three halves. Zero plus one is 11, times negative three halves is negative three halves plus positive three halves is the zero that we want. Ye and three times negative three halves is negative nine halves plus 11 halves simplifies to one. Look at this, this has our diagonal row of ones. And therefore we recall from the beginning of this lesson that this is our corresponding A. B. And C. And we remember that we were trying to fill that in for we had A X. Squared. So that's a negative two X. Squared plus B. X. Well that's A plus one X. Or plus X. And then plus C. Well that's plus three are corresponding A. B. And C. Terms equal to that F. Of X. And that's our answer. That's what we are looking for. Now let's scroll back up and see what matches mess for a second fit this in here. So now we look at our answer choices and this matches with answer choice. B. Well done. We'll catch you on the next one.

Find the quadratic function f(x) = ax² + bx + c for which ƒ( − 2) = −4, ƒ(1) = 2, and f(2) = 0.

Key Concepts

Quadratic Functions

Function Evaluation

Solving Systems of Equations

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