SOLUTION: 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 a
Algebra.Com
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....
RELATED QUESTIONS
The equation of a curve is 3 4 2
y x x= + − .
i Show that the equation of the normal... (answered by ikleyn)
The equation of a curve is y = x^2/(x+2). The tangent to the curve at the point where x = (answered by MathLover1)
Find the equation of the line normal to the curve {{{y^2=9-24/(6-x)}}}
at the point... (answered by Edwin McCravy)
Please help me solve this equation:
1. Determine dy/dx for y = x√x(5x^2 − (answered by Fombitz)
Find the equation of the normal to the curve y = –x^2 + 4x at:... (answered by solver91311)
A curve has the equation {{{y = (ax+3)ln(x)}}}, where x is greater than zero and a is a... (answered by Tatiana_Stebko)
Find the equation of the normal to the curve y = –x2 + 4x at:... (answered by josgarithmetic)
Determine the equation of the normal line at the point x=4 to the curve... (answered by Alan3354)
Find the equation of the normal and tangent to the curve at the given point:
f(x) =... (answered by Alan3354)