Question 1127558: The equation of a curve is y=3+4x-x^2.
Show that the equation of the normal to the curve at a point (3,6) is 2y=x+9.
Given that the normal meets the coordinate axes at points A and B ,find the coordinates of midpoint of AB.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The function is y = -x^2+4x+3
The derivative is -2x+4
The derivative at x=3 is -2(3)+4 = -2; so the slope of the curve at x=3 is -2.
The slope of the normal to the curve at x=3 is the negative reciprocal, 1/2.
The equation of the line with slope 1/2 passing through (3,6) is y = (1/2)x+9/2, or 2y = x+9.
The x-intercept of the normal (set y=0) is (-9,0); the y-intercept (set x=0) is (0,9/2).
The midpoint of the segment with endpoints (-9,0) and (0,9/2) is (-9/2,9/4).
A graph....
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