Answer Check - Find an equation of the tangent line

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SUMMARY

The discussion focuses on finding the equation of the tangent line to the curve defined by the equation y = 5x³ at the point (-3, -135). The derivative of the curve is calculated as y' = 15x², and at x = -3, the slope is confirmed to be 135. To find the equation of the tangent line, one must use the point-slope form of a line, which results in the equation y + 135 = 135(x + 3).

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Homework Statement


Find an equation of the tangent line to the curve y = 5x3 at the point

(-3,-135). Enter your equation in y = form.

Homework Equations




The Attempt at a Solution



Taking the derivative of the equation I get

y = 15x2, following that I plug in the x value of the given point:
y = 15(-3)2
and get
y = 135

Am I doing this correctly?
 
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First, you should write the derivative of y with a different symbol than "y". Conventional notations for the derivative are y' or dy/dx.

So you have found

y'(x) = 15x² ==> y'(-3)=135,

which is correct. But y'(3) represents the slope of the curve y(x) at the point (-3,-135). You are asked to find an equation for the line of slope 135 and passing through the point (-3,-135).
 

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