For what values of a and b is the line -4x+y=b tangent to the curve y=ax^3 when x=4?

Can anyone give me the hint?

2 answers

dy/dx = 3 ax^2
when x = 4, dy/dx= 48 a = slope

but y = 4 x + b has slope of 4
so
48 a = 4
a = 4/48 = 1/12
and
y = (1/12) x^3
when x = 4
y = 16/3

-4(4) + 16/3 = b
-16 + 16/3 = b

b = -32/3

so
you can verify this at

http://www.wolframalpha.com/input/?i=plot+y%3D1%2F12+x^3,+y%3D4x-32%2F3,+0%3C%3Dx%3C%3D6
Similar Questions
    1. answers icon 1 answer
  1. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 1 answer
  2. Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 .a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1) b. Write an equation of each
    1. answers icon 3 answers
  3. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 0 answers
more similar questions