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) .

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Expert Answer

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)

p(x) axbx2+cx+d
P(1) a(1)+b(1)c(1)+d
1ab+c+d
P(-1) a(1b(-1)2+c(-1)+d
-3=-a+b-c+d
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p(x) axbx2+cx+d P(1) a(1)+b(1)c(1)+d 1ab+c+d P(-1) a(1b(-1)2+c(-1)+d -3=-a+b-c+d

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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)

 

+bx2+cx+d
у-ах
y' 3ax2+2bx+c
14-3a(1)2+2b(1)+c|
За+2b+с-14
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+bx2+cx+d у-ах y' 3ax2+2bx+c 14-3a(1)2+2b(1)+c| За+2b+с-14

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Step 3

Slope of y = y = - 2x - 5  is = -2

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

у-ах' +bx?+сх+d
y' 3ax2+2bx+c
-2=3a(-1)+2b(-1)+c
3a-2b+c-2
help_outline

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у-ах' +bx?+сх+d y' 3ax2+2bx+c -2=3a(-1)+2b(-1)+c 3a-2b+c-2

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Math

Calculus