SOLUTION: Suppose that A, B, C, and D are constants and f is the cubic polynomial f(x)=Ax³+Bx²+Cx+D. Suppose also that the tangent line to y = f(x) at x = 0 is y = x and the tangent line at

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Suppose that A, B, C, and D are constants and f is the cubic polynomial f(x)=Ax³+Bx²+Cx+D. Suppose also that the tangent line to y = f(x) at x = 0 is y = x and the tangent line at       Log On


   

Question 350591: Suppose that A, B, C, and D are constants and f is the cubic polynomial f(x)=Ax³+Bx²+Cx+D. Suppose also that the tangent line to y = f(x) at x = 0 is y = x and the tangent line at x = 2 is given by y = 2x – 3. Find the values of A, B, C, and D. Then sketch the graph of y = f(x) and the two tangent lines for -2 ≤ x ≤ 4.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
y=Ax%5E3%2BBx%5E2%2BCx%2BD
The slope of the tangent line at a given point is the value of the derivative at that point.
The derivative of the function is,
dy%2Fdx=3Ax%5E2%2B2Bx%2BC
At x=0, the slope of the tangent line is m=1.
m=dy%2Fdx%280%29=3A%280%29%2B2B%280%29%2BC=1
highlight%28C=1%29
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At x=2, the slope of the tangent line is m=2.
m=dy%2Fdx%282%29=3A%282%29%5E2%2B2B%282%29%2B1=2
12A%2B4B%2B1=2
1.12A%2B4B=1
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Using the tangent line at x=0, you also know the intersection point because,
y=x
y=0
(0,0) is also a point on the cubic.
y=Ax%5E3%2BBx%5E2%2BCx%2BD
y=A%280%29%2BB%280%29%2B%280%29%2BD=0
highlight%28D=0%29
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Similarly at x=2,
y=2x-3
y=2%282%29-3
y=1
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y=Ax%5E3%2BBx%5E2%2Bx
1=A%282%29%5E3%2BB%282%29%5E2%2B2
1=8A%2B4B%2B2
2.8A%2B4B=-1
Subtract eq. 2 from eq. 1,
12A%2B4B-8A-4B=1-%28-1%29
4A=2
highlight%28A=1%2F2%29
Then from eq. 2,
8%281%2F2%29%2B4B=-1
4%2B4B=-1
4B=-5
highlight%28B=-5%2F4%29
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