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

y=3x(x+1)2,  (4,1225)y=3x(x+1)2,  (4,1225)



y=y= 

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

 

y=2x+1,  (4,3)y=2x+1,  (4,3)

 

y=

Consider the slope of the given curve at each of the five points shown. List these five points in decreasing order of the slopes.

On a separate piece of paper, sketch the graph of the parabola y=x2+3y=x2+3. On the same graph, plot the point (0,4)(0,−4). Note that there are two tangent lines of y=x2+3y=x2+3 that pass through the point (0,4)(0,−4).

Specifically, the tangent line of the parabola y=x2+3y=x2+3 at the point (a,a2+3)(a,a2+3) passes through the point (0,4)(0,−4) where a>0a>0. The other tangent line that passes through the point (0,4)(0,−4) occurs at the point (a,a2+3)(−a,a2+3).

Find the number aa. 

Let f(x)={xsin10x0 if x0 if x=0.f(x)={xsin⁡−10x if x≠00 if x=0.
Determine whether or not f(0)f′(0) exists.

Answer Yes or No:

 

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