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Find a cubic function $ y = ax^3 + bx^2 + cx + d $ whose graph has horizontal tangent at the points (-2, 6) and (2, 0).

$$y=\frac{3}{16} x^{3}-\frac{9}{4} x+3$$

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he had slower. So when you right here. So we have. Why is equal to a X cute? Must be X square plus feet x plus D you differentiate this to get three a X Square plus two B X? Let's see. We're gonna set up our equations from our given points, so we have negative to calm a six when we plug it into the original equation. We got six in our line ISS negative. Eat a That's four p minus two C plus T is equal to six. Then we have 0.2 comma zero. When we plug it into our original equation, we good cereal. Yeah, a Ame plus four B plus two C plus D is equal to zero. Where are derivatives since there's horizontal tangents, the derivative for positive and negative, too, is equal to zero, so we go for negative to calm a six. You get our derivative to be equal to zero well, a minus four B plus C. It's equal to zero and for our point to calm a zero. This is also there as well, since there are horizontal tangents. 12 a plus four B plus C is equal to zero then we're going to subtract our equations. 12 day was for P. Thus he is equal to zero minus 12. A finest for P plus C is equal to zero. Then we got B is equal to zero. We're gonna add up our equations. Our 1st 2 equations less for B plus two c rusty this equal to zero. Then we got Dee as equal to three. Since we plug in, zero for B C is equal to negative 12 A. When we get a was four times zero plus two negative 12 a plus three is equal to zero. Can we get a value of three over 16? And when we plug it in, C becomes negative nine over four. Then we just plug it in to get wise people to three over 16 x square x cubed Excuse me minus nine over four x plus three