How to find a equation that is tangent to f(x)=1/sqrtxf(x)=1√x and parallel to the line -(1/2)x+3−(12)x+3?
1 Answer
Aug 21, 2017
Please see below.
Explanation:
The slope of the tangent at a point is the value of the first derivative at that point i.e. if we are seeking a tangent at
Here given function is
As tangent is parallel to line
Hence tangent is
graph{(y-1/sqrtx)(2y+x-6)(x+2y-3)=0 [-1.275, 8.725, -1.15, 3.85]}