Find the coordinates of the turning point when y=5x^3 - 6x^2 - 4X + 8

Question

asked Jun 14, 2022 210k views

1 Answer

James Ralston · Answer 1 · 2022-06-20T10:18:39+0000

Given:

The equation is

y=5x^3-6x^2-4x+8

To find:

The coordinates of the turning point.

Solution:

We have,

y=5x^3-6x^2-4x+8

Differentiate with respect to x.

(dy)/(dx)=5(3x^2)-6(2x)-4(1)+(0)

(dy)/(dx)=15x^2-12x-4

For the turning points
(dy)/(dx)=0 .

15x^2-12x-4=0

Using quadratic formula:

$x=(-b\pm √(b^2-4ac))/(2a)$

$x=(-(-12)\pm √((-12)^2-4(15)(-4)))/(2(15))$

$x=(12\pm √(144+240))/(30)$

$x=(12\pm √(384))/(30)$

Now,

$x=(12+√(384))/(30)\text{ and }x=(12-√(384))/(30)$

$x\approx 1.053\text{ and }x\approx-0.253$

Putting x=1.053 in the given equation, we get

y=5(1.053)^3-6(1.053)^2-4(1.053)+8

$y\approx 2.973$

Putting x=−0.253 in the given equation, we get

y=5(-0.253)^3-6(-0.253)^2-4(-0.253)+8

$y\approx 8.547$

Therefore, the turning points are (1.053,2.973) and (-0.253, 8.547).