Question

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

Answer 1

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.

$\dfrac{dy}{dx}=5(3x^2)-6(2x)-4(1)+(0)$

$\dfrac{dy}{dx}=15x^2-12x-4$

For the turning points $\dfrac{dy}{dx}=0$.

$15x^2-12x-4=0$

Using quadratic formula:

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-(-12)\pm \sqrt{(-12)^2-4(15)(-4)}}{2(15)}$

$x=\dfrac{12\pm \sqrt{144+240}}{30}$

$x=\dfrac{12\pm \sqrt{384}}{30}$

Now,

$x=\dfrac{12+\sqrt{384}}{30}\text{ and }x=\dfrac{12-\sqrt{384}}{30}$

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

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).