The point P(1,-2) lies on the curve y=x^3-3x.
a) if Q is the point (x,x^3-3x), find the slope of the secant line PQ for x=2
b) FInd the slope of the tangent line to the curve at P.
c) Find an equation of the tangent line to curve of P.
d) Graph the curve and the tangent line.
Thanks!

Question

The point P(1,-2) lies on the curve y=x^3-3x.
a) if Q is the point (x,x^3-3x), find the slope of the secant line PQ for x=2 
b) FInd the slope of the tangent line to the curve at P.
c) Find an equation of the tangent line to curve of P.
d) Graph the curve and the tangent line.
Thanks!

anonymous · Answer

i figured out part a), which is m=4

freckles · Answer

Hey, do you know the short cut derivative ways?

anonymous · Answer

sadly, no

freckles · Answer

like power rule constant multiple rule ...

freckles · Answer

Ok but you do know the algebraic definition of derivatie

freckles · Answer

$$(x^3-3x)'=\lim_{h \rightarrow 0}\frac{((x+h)^3-3(x+h))-(x^3-3x)}{h}$$
Can you evaluate this limit?

freckles · Answer

Hint: play with the top a lot...
(x+h)^3 <--multiply this out 
And combine any like terms on top!

freckles · Answer

Do just the work to the top and let me know what you get

freckles · Answer

@esam2 Have you multiplied the (x+h)^3 out?

freckles · Answer

You can use pascal's triangle

freckles · Answer

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