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Social ScienceFinding 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) .
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) .
Expert Answer
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)
Image Transcriptionclose
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
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)
Image Transcriptionclose
+bx2+cx+d у-ах y' 3ax2+2bx+c 14-3a(1)2+2b(1)+c| За+2b+с-14
Slope of y = y = - 2x - 5 is = -2
So slope of y=ax^3 + bx^2 + cx + d a...
Image Transcriptionclose
у-ах' +bx?+сх+d y' 3ax2+2bx+c -2=3a(-1)+2b(-1)+c 3a-2b+c-2
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Calculus
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