Question

Asked Nov 3, 2019

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Finding a Polynomial Find a third degree polynomia p(x) that is tangent to the line y = 14x - 13 at the point (1, 1) . and tangent to the line y = - 2x - 5y the point (- 1, - 3) .

Step 1

Let p(x) = ax^3 + bx^2 + cx + d

This passes through (1,1) and (-1,-3) .

For (1,1), x=1 and y=1. For x=(-1,-3) , x = -1 and y =-3.

Plugging them we get two equations of a,b,c,d.

a+b+c+d=1 -----(Eq. 1)

-a+b-c+d=-3 ------(Eq. 2)

Step 2

Slope of y = 14x - 13 is = 14

So slope of y=ax^3 + bx^2 + cx + d at (1,1) is 14.

From there we get 3a+2b+c=14 ......(Eq. 3)

Step 3

Slope of y = y = - 2x - 5 is = -2

So slope of y=ax^3 + bx^2 + cx + d a...

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