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Find an expression for a cubic function $ f $ if $ f(1) = 6 $ and $ f(-1) = f(0) = f(2) = 0 $.

$-3 x^{3}+3 x^{2}+6 x$

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for this problem. We're looking for the equation of a cubic function, and we know that it goes through these four points and we could write this equation in general as f of X equals a X cubed Class B X squared plus C X plus D. And our goal is to find the values of ABC. Indeed. So what we can do is substitute these ordered pairs in one by one, and we'll have a system with four equations and for unknowns. So I'm going to start by using the 0.16 substituting six in for F of X and one in for X and I get six equals a times one cubed plus B times one squared plus C times one plus de. We can simplify that in its six equals a plus B plus C plus d. Okay, let's go through that process for the other points. Now we're looking at the point negative 10 So we substitute zero in for f of X and negative. One in for X zero equals a times negative one cubed plus B times negative one squared plus C times negative one plus de, and that gives us zero equals the opposite of a plus B minus C plus D. Okay, let's move on to the third point, which was 00 We substitute zero in for X and zero, and for F of X, we get zero equals a time. Zero. That's just zero plus B time. Zero. That's just zero plus C time. Zero. That's zero plus d. So now we're getting somewhere. We know that d is zero. That's a relief. Okay, we have one more point to go through this process with, and that's the 10.0.20 So we use that point. We have zero equals. And just to save on a little space, I'll just talk through some without writing it. So we have a times two cubed and that's going to be eight A. We have be times two squared, and that's going to be four b. We have C times, too, so that's to see and we have d. But we just found out that d a zero and that was supposed to be a C right there. Okay, let's look at what we have now. We can go back to our first and second equations, and remember that D is just zero. So let's just eliminate the D. So we have six equals a plus. B plus C zero equals the opposite of a plus B plus C and zero equals eight a plus four B plus two seat. We have a system of three equations, three unknowns. So now we want to think about how to solve that system. So here we have the system again and a little bit more space to work, and I'm going to label these equations 12 and three. Now we have a lot of different options for how to solve a system of three equations. Do you mind even know some matrix options like Cramer's rule or the inverse Matrix method? And you might be able to do those on a graphing calculator? And that would be great. What I'm going to do here is just a straightforward elimination method, and I'm going to take equations one and two, and I'm going to add them together, and that is going to eliminate variable a and variable See. So if six equals a plus, B plus C and we have zero equals the opposite of a plus B minus C. And when we add those together we end up with six equals to be so if we divide that by two, we get three equals b. Okay, so now we know B and we already found the earlier. Now we need to find A and C. So let's go back to equation too. And let's substitute RB value in there. The equation, too. We have zero equals the opposite of a plus three minus C and let's go to Equation three. And let's substitute RB value in there and we have zero equals eight a plus four times B. So that would be plus 12 plus to see. Okay, Now let's work with this system. So if we were to multiply the second equation Equation two were to multiply that by two, we could add these together and we would eliminate C. So I'm going to take Equation two and double it zero equals negative to a plus six minus two c. I'm going to leave Equation three as it is. Zero equals eight A plus 12 plus to see. I'm going to add these equations together to eliminate C so we have zero equals six A plus 18. We can solve that for a subtract 18 from both sides and divide by six. And we get a equals negative three. Okay, we only have one more value to find, and that would be See. So let's go back to equation one. We have six equals a plus. B plus C A is negative. Three b is three and we're gonna find See, this gives us six equal C c equal six. Okay, Now we have left To do is write thes into our equation of the cubic function. So we have f of x equals a so negative three times x cubed plus B so three times x squared Class C, six times x plus de zero and there's the function.