# Analyze and sketch a graph of the function. Label any

Analyze and sketch a graph of the function. Label any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to verify your results $y=3{x}^{\frac{2}{3}}-2x$
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Luis Sullivan

Step 1 Given Function
$y=3{x}^{\frac{2}{3}}-2x$
To Find: Relative extreme, Point of inflection and Asymptotes.
On Sketching the graph of given function we get,

Step 2: From first derivative test definition
Suppose that isa critical point of then
If $f‘\left(x\right)>0$ to the left of $x=c$ and $f‘\left(x\right)<0$ to the right of $x=c$ then $x=c$ is a local maximum.
If $f‘\left(x\right)<0$ to the left of $x=c$ and f(x)>0 to the right of $x=c$ then $x=c$ is a local minimum.
If f(x) is the same sign on both sides of $x=c$ then $x=c$ is neither a local maximum nor a local local minimum.
Step 3: On differentiating the given equation we obtain,
$\frac{dy}{dx}=3×\left\{\frac{2}{3}\right\}{x}^{-\frac{1}{3}}-2$
$\frac{dy}{dx}=\left\{\frac{2}{{x}^{\frac{1}{3}}}\right\}-2$
Now, to find critical points substitute,
$\frac{dy}{dx}=0$
$\left\{\frac{2}{{x}^{\frac{1}{3}}}\right\}-2=0$
$\left\{\frac{2}{{x}^{\frac{1}{3}}}\right\}=2$
${x}^{\frac{1}{3}}=1$
$x=1$
So the critical points obtained
$x=O$ and $x=1$
$-\propto$ So the intervals are
Checking the sign of f`(x)
at each monotone interval we have,
Step 5: By the Inflection point Definition
An inflection point is a point on graph at which the second derivative changes sign
If $f\left(x\right)>0$ then f(x) concave upwards
IF $f\left(x\right)<0$ then f(x) concave downwards
Here, We have,
$f\left(x\right)=-\frac{2}{3{x}^{\frac{4}{3}}}$
Checking the sign we obtain,
$\begin{array}{|cccc|}\hline & -\propto

On the Above analysis we find that there are
No any point of Inflection that we have for the given function.
Resulting in No any Vertical as well as Horizontal Asymptotes.