What is the end behavior and turning points of #y = -2x^3 + 3x - 1#?

Question

James Allen · Accepted Answer

As #x->oo, y->-oo and as x-> -oo, y->oo#

Turning points are #x= -1/sqrt2, +1/sqrt 2#.

For end behaviour, note the leading coefficient and the degree. The degree is odd and the leading coefficient negative, hence it would rise to the left and fall to the right. For turning points get y'=0 and solve. In this case it is #-6x^2+3=0# This gives #x= -1/sqrt2, +1/sqrt 2#.

Sadie Alexander · Answer

To find the turning points, differentiate the function and set it to zero.
This yields : #dy/dx=0iff -6x^2+3=0 iff x = +-1/4# The corresponding y-values are then #y(-1/4)=-55/32# and #y(1/4)=-9/32# To find the regions in which the function is increasing and decreasing, we investigate the sign of the first derivative around the turning points and to find the inflection point we set the second derivative to zero and find x = 0 and the corresponding y values is then #y(0)=-1#
To find the concavity, we investigate the sign of the second derivative on each side of the inflection point. Since #y(1)=0=>x=1 # is a root and x intercept and so #(x-1) # is a factor. We may then long divide to find the other factors and roots. Putting all together, we get the graph :   graph{-2x^3+3x-1 [-10, 10, -5, 5]}

Theodore Amidon · Answer

undefined The end behavior of the polynomial function $ y = -2x^3 + 3x - 1 $ is determined by the leading term, which is $ -2x^3 $. As $ x $ approaches positive or negative infinity, the leading term $ -2x^3 $ dominates the behavior of the function. Since the coefficient of the leading term is negative, the end behavior is as follows:

- As $ x $ approaches positive infinity, $ y $ approaches negative infinity.
- As $ x $ approaches negative infinity, $ y $ approaches negative infinity as well.

The turning points of the function correspond to the points where the derivative of the function $ y' $ equals zero. To find the turning points, we first find the derivative of $ y $ with respect to $ x $, and then solve for $ x $ when $ y' = 0 $. After finding these $ x $-values, we can evaluate $ y $ at these points to find the corresponding $ y $-values.

$$ y' = \frac{d}{dx}(-2x^3 + 3x - 1) = -6x^2 + 3 $$

Setting $ y' = 0 $ to find critical points:

$$ -6x^2 + 3 = 0 $$
$$ -6x^2 = -3 $$
$$ x^2 = \frac{1}{2} $$
$$ x = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2} $$

These critical points correspond to the potential turning points of the function. To find the corresponding $ y $-values, we substitute these $ x $-values back into the original function:

$$ y\left(\frac{\sqrt{2}}{2}\right) = -2\left(\frac{\sqrt{2}}{2}\right)^3 + 3\left(\frac{\sqrt{2}}{2}\right) - 1 $$
$$ y\left(\frac{\sqrt{2}}{2}\right) = -2\left(\frac{1}{2\sqrt{2}}\right) + \frac{3\sqrt{2}}{2} - 1 $$
$$ y\left(\frac{\sqrt{2}}{2}\right) = -\frac{1}{\sqrt{2}} + \frac{3\sqrt{2}}{2} - 1 $$
$$ y\left(\frac{\sqrt{2}}{2}\right) = \frac{3\sqrt{2}}{2} - \frac{1}{\sqrt{2}} - 1 $$
$$ y\left(\frac{\sqrt{2}}{2}\right) = \frac{3\sqrt{2}}{2} - \frac{\sqrt{2}}{2} - 1 $$
$$ y\left(\frac{\sqrt{2}}{2}\right) = \frac{2\sqrt{2}}{2} - 1 $$
$$ y\left(\frac{\sqrt{2}}{2}\right) = \sqrt{2} - 1 $$

Similarly,

$$ y\left(-\frac{\sqrt{2}}{2}\right) = -\sqrt{2} - 1 $$

Therefore, the turning points are approximately $ \left(\frac{\sqrt{2}}{2}, \sqrt{2} - 1\right) $ and $ \left(-\frac{\sqrt{2}}{2}, -\sqrt{2} - 1\right) $.