Question

Find the slope of the functions graph at a given point. Then find an equation for the line tangent to the graph there.

f(x)=2*square root of x. (1,2)

Could someone please help me figure this one out?

Answer 1

given
f(x)=2√x
Therefore f(1)=2*√1=2, which means that (x0,y0)=(1,2) is a point on the curve (given).

Find f'(x)=(d/dx)2√x=1/√x
the tangent at (1,2) is the line passing through (1,2) with a slope equal to it's slope, namely f'(1).

Thus the required line is:
L : (y-y0)=f'(1)*(x-x0)
or
L : (y-2)=f'(1)*(x-1)

I will let you evaluate f'(1) to get the final solution.