y = 3x^2 - x - 5
y' = 6x-1
So, at (1,-3), the slope of the normal is -1/5, and the line there is
y+3 = -1/5 (x-1)
y = -1/5 x - 14/5
So, where does the line intersect the parabola?
-1/5 x - 14/5 = 3x^2-x-5
-x-14 = 15x^2-5x-25
15x^2-4x-11 = 0
(15x+11)(x-1) = 0
So, the line intersects at x=1 and x = -11/15
Now you just need y at x = -11/15
At what point does the normal to
y=-5-1x+3x^2 at (1, -3 ) intersect the parabola a second time?
Answer:
Note: You should enter a cartesian coordinate.
The normal line is perpendicular to the tangent line. If two lines are perpendicular their slopes are negative reciprocals -- i.e. if the slope of the first line is m then the slope of the second line is - 1/m
1 answer
1 answer