A curve c has equation x^3-x^2-8x+12= y

Question

1. A curve c has equation x^3-x^2-8x+12= y

a- find dx/dy

b- the curve c has 2 turning points

work out the x coordinates of the two turning points

c- show that the c axis is a tangent to the curve c

Answer 1

x^3-x^2 -8x +12= y

y'(x) = dy/dx = 3x^2 -2x -8

set it =0 to find extrema, local max and minimums

3x^2 -2x -8 =0

x = 2/6 +/-(1/6)sqr(4+96)

= 1/3 +/-10/6 = -4/3 or 2

x = 2 or -4/3

y(2) = 0

y(-4/3) =(-4/3)^3 -(-4/3)^2 - 8(-4/3) +12

= -64/27 +16/9 +32/3 +12

=(-64 +64 +288 +324)/27

= (288+324)/27

=612/27 = 68/3 = 22 2/3

turning points are (2,0) and (-4/3, 68/3)

to the left of (2,0) the curve goes downhill

between (2,0) and (- 1 1/3, 22 2/3) the curve goes up hill

then to the right of (-1 1/3, 22 2/3) the curve goes down hill approaching negative infinity

1st turning point (2,0) is a local or relative minimum point of the curve

2nd turning point (-1 1/3, 22 2/3) is a local or relative maximum point

there are no global max or minimums

at the point (2,0) the tangent to the curve is the x axis

use a graphing calculator to see it all better

if you're really looking for dx/dy it = x'(y) = the inverse of dy/dx = 1/dy/dx = 1/(3x^2-2x-8)