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Please..I'm desperate. I have a D in this class..

Find an equation of the tangent line to the curve at the given point. y = 6x - x^3, (1,5)​

User Andy Xu
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1 Answer

4 votes

Explanation:

Recall that the first derivative of an equation gives us the SLOPE for the tangent line.

So first, we should find the first derivative.


(dy)/(dx) = 6 - 3x^2

With the first derivative, we can solve for the SLOPE of the tangent line at (1,5).

Using x = 1.


slope = 6 - 3(1)^2

A tangent line is a line. And has the form
y = mx +b

Where m is the SLOPE of the line. and b is the y-intercept.

To get to this equation, let's use the point-slope technique.


y-y_1 = m(x-x_1)


y - 5 = m(x-1)

Solve for slope, substitute it into this equation for m and solve. That is the equation of the tangent line at the point (1,5).

User Rajat Modi
by
6.5k points
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