Find the equation of the tangent line to the curve at the given point?

Question

I am trying to find the equation of the tangent line to the curve (y = x^3 - 3x + 1) at the given point (3, 19). I have already calculated the derivative, which is (f'(x) = 3x^2 - 3). However, I'm not sure how to proceed from there. Could someone guide me through the next steps to solve this problem? Thank you!

Anonymous · Accepted Answer

Okay, since you already know the derivative, you found the equation of the slope. So you take the derivative and plug in the given x value to find the value of the slope.
y'=3x^2-3
y'=3(3)^2-3 which equals 24
Now since you know what the value of the slope is (m=slope) and you have a x and y value (3,19), then you can plug it into the slope intercept form to find the value of b. The equation is y=mx+b.
19=(24)(3)+b
19=72+b
b=-53
Now since you have the value of b and m, you just put it into y=mx+b. 
y=24x-53 is the equation of the tangent line.

Anonymous · Answer

y = x³ - 3x + 1
y´ = 3x² - 3
What is the derivative? It gives us the slope of the original function right?
So if we want a line tangent to the original, we had better know the slope at that point.
f´(3) = 3(3)² - 3
= 24
So a line's equation is y=m*x + b right? well we found m and we know a (x,y) so lets find b
(19) = 24(3) +b
=-53
Your tangent line is: y = 24x -53

Anonymous · Answer

f'(3) = slope of line
(3,19) = point on line
use point-slope formula

Anonymous

Find the equation of the tangent line to the curve at the given point?

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