Question

find the points on the graph of y=x^3+2x^2-4x+1

at which tangent line is perpendicular to 4y-x=3

4y−x=3

Answer 1

The line4y -x =3 rearranges as y=x/4+3/4, so the slope is 1/4. Any perpendicular line would have a slope of -4.

y = x^3+2x^2 -4x +1

y' = 3x^2 + 4x - 4

So if y' = -4

3x^2 + 4x - 4 = -4

3x^2 + 4x - 4+ 4 = 0

x(3x+4) = 0

x =0, -4/3

So the tangents perpenducular to the given line go through (0,1) and (-4/3, 203/27).

y -1= - 4(x - 0)

y -1 = -4x

y = -4x+1

y - 203/27 = - 4(x+4/3)

y = -4x - 16/3 + 203/27

y = -4x + (-144+203)/27

y = - 4x + 59/27

Graph of the