How do you find the equation of the line tangent to the graph of #y sqrt(x + 1) = 4# at the point x=3?

Question

2977 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-21T01:46:42

See the explanation section, below.

#y sqrt(x + 1) = 4#

#y = 4/sqrt(x+1) = 4(x+1)^(-1/2)#.

The point on the graph where #x=3# has #y# coordinate #4/sqrt(3+1) =2#.

So the line contains the point #(3,2)#

So, #dy/dx = -2(x+1)^(-3/2) = (-2)/sqrt(x+1)^3#

To get the slope of the line tangent to the graph at the point where #x=3#, evaluate #dy/dx# at #x=3#.

#m = (-2)/sqrt(3+1)^3 = (-2)/8 = -1/4#

The equation of the line through #(3,2)# with slope #m=-1/4# is

#y=-1/4x+11/4#