Asked by Hanna
The point on the curve for y = sqrt(2x+1) at which tangent is perpendicular to the line y = -3x + 6 is....
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?
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?
Answers
Answered by
Steve
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)
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)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.