How can the turning point on the graph be estimated

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.

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

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

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.

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