0
$\begingroup$

If we know for a fact that the turning point of a graph is not always exactly between the two $x$-intercepts, I just came across the following

$y=(x+2)^3(x-2)^2$

I can use a CAS calculator, but without one, how do I know if the turning point is supposed to be before or after the $y$-axis (I mean the $y$-intercept in this case) if I know that the $x$-intercepts are equidistant from the origin, like $2$ and $-2$

Note that the turning point even in this case is not exactly between the two intercepts even the intercepts themselves are equidistant from the origin.


EDIT:

I think the question is not clear enough. By now I am not even required to find the turning point when sketching, the basics needed are the intercepts and end end behaviour. However, I need an estimate for the turning point, something very rough, like below or above x-axis, just so that the basic shape of the graph is identified.

Again, I do not need to find the turning point, just guess something like in which quadrant would it lie.

$\endgroup$
  • $\begingroup$ Plot $$f'(x)$$ and calculate the extrempoints of $$f'(x)$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 25 '18 at 12:54
0
$\begingroup$

$$y=(x+2)^3(x-2)^2$$ Use the product rule to achieve $$\frac{dy}{dx}=2(x+2)^3(x-2)+3(x+2)^2(x-2)^2$$ Plot the graph and find where $$\frac{dy}{dx}=0$$Here $x=\frac{2}{5}, x=\pm 2$

You can then plug in those x values to find the y values so: $$(2,0); (-2,0); (\frac{2}{5}, \frac{110592}{3125})$$ If you want to know whether they are a minimum or maximum, differentiate again to get $\frac{d^2y}{dx^2}$, plug in the solutions to $\frac{dy}{dx}=0$, if the result is positive, your curve is on its way up hence the turning point is a minimum, if it's negative, the turning point is a maximum.

$\endgroup$
0
$\begingroup$

Turning points are where the derivative changes sign. $$ y=(x+2)^3(x-2)^2 \implies y'=(x-2)(x+2)^2(5x-2) $$

$$(x-2)(x+2)^2(5x-2)=0 \implies x=2,-2,2/5 $$

$(x+2)^2$ does not cause any change in sign of derivative.

Turning points are at $x=2$ and $x=2/5.$

It all depends on how the derivative behaves.

$\endgroup$
  • $\begingroup$ The turning point at $x=-2$ is called a Point of Inflection if I recall correctly and can be called a turning point $\endgroup$ – Rhys Hughes Apr 27 '18 at 21:02
  • $\begingroup$ yes, $x=-2$ is an inflection point. where the concavity changes. It is a point of diminishing return. $\endgroup$ – Mohammad Riazi-Kermani Apr 27 '18 at 23:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.