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Perpendicular tangent trinomials?
A. Find an equation for the line perpendicular to the tangent line to the curve y=x^3-16x+3 at the point (4,3).
b. What is the smallest slope on the curve? At what point on the curve does the curve have this slope?
c. Find equations for the tangent lines to the curve at the points where the slope of the curve is 32.
image: https://imgur.com/3og48GO
2 Answers
- Wayne DeguManLv 79 months agoFavorite Answer
dy/dx = 3x² - 16
At x = 4 we have, dy/dx = 3(4)² - 16
i.e. 32
The perpendicular will have gradient -1/32
Hence, y - 3 = (-1/32)(x - 4)
so, 32y - 96 = 4 - x
i.e. 32y + x - 100 = 0
As dy/dx is a parabola with vertex at (0, -16), the smallest slope is -16.
This happens at x = 0 => y = 3...i.e. at (0, 3)
When dy/dx = 32 we have:
3x² - 16 = 32
so, x² = 16
i.e. at x = -4 and at x = 4
Hence, at points (-4, 3) and (4, 3)
so, y - 3 = 32(x + 4) and y - 3 = 32(x - 4)
i.e. y = 32x + 131 and y = 32x - 125
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