The point on the curve for y = sqrt(2x+1) at which tangent is perpendicular to the line y = -3x + 6 is....

Question

a. (4, 3) b. (0,1) c. (1, sqrt3) d. (4, -3) e. (2, sqrt 5)

I'm not sure, but do you find the perpendicular slope (1/3) set that as dy/dx and solve for x or y? I tried that and it didn't work. Help, please?

Answer 1

the slope of the tangent at any point (x,y) is dy/dx = 1/sqrt(2x+1)

The slope of the line given is -3
The slope of the perpendicular to that line is thus 1/3

So, you want 1/3 = 1/sqrt(2x+1)
3 = sqrt(2x+1)
9 = 2x+1
x = 4
y(4) = 3

so, the point in question is (4,3)