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

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y = x^3 - 3x + 1 Given Point: (3,19)I just cannot figure out how to solve this problem. I found the derivative at f'(x) = 3x^2 - 3, but I do not know where to go from there.

Dr. Juanita Wolff · 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^2-3y'=3(3)^2-3 which equals 24Now 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)+b19=72+bb=-53Now 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.

Dr. Juanita Wolff · Answer

y = x³ - 3x + 1y´ = 3x² - 3What 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= 24So 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=-53Your tangent line is: y = 24x -53

Dr. Juanita Wolff · Answer

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

Dr. Katlynn Swaniawski II

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

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