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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

Hello,

I wanted to take some time and show how to sketch the graph of a function! If you see anything that is wrong, please PM me and I will correct it! I hope you enjoy, understand and learn!

1. An exponential function

Draw the graph of \(\displaystyle y=e^{-2x^3+3x^2+1}\)

Okay, the first thing I notice is that the graph will never cut the $x$-axis because for any real $x$:

\(\displaystyle e^{-2x^3+3x^2+1}\neq0\)

If we look at y intercept ( Where the function cut the \(\displaystyle y\)-axis that means \(\displaystyle x=0\) so we get

\(\displaystyle e^{-2*0^3+3*0^2+1}=e\)

Okay, next we want to differentiate with respect to $x$ because we want to look at the slope.

So how do we differentiate that function?

Using the chain rule, we find:

\(\displaystyle \frac{dy}{dx}=e^{-2x^3+3x^2+1}\frac{d}{dx}(-2x^3+3x^2+1)=e^{-2x^3+3x^2+1}(-6x^2+6x)=6x(1-x)e^{-2x^3+3x^2+1}\)

If we want to find the critical values, we have to equate the derivative to zero (this is where the slope is zero)

Because \(\displaystyle e^{-2x^3+3x^2+1}\neq0\) this leaves us with:

\(\displaystyle 6x(1-x)=0\)

If we solve that equation we get \(\displaystyle x=1\) and \(\displaystyle x=0\) and it is at those $x$-values we will have extrema since they are roots of odd multiplicity and we therefore know the sign of the derivative will change across these critical values.

Let's make a schedule to analyze the intervals of increasing/decreasing behavior, and we see:

So, we may conclude that the given function is:

decreasing on \(\displaystyle (-\infty,0)\)

increasing on \(\displaystyle (0,1)\)

decreasing on \(\displaystyle (1,\infty)\)

Next, let's find the asymptotes.

Note: Only rational functions have oblique or slant asymptotes, so there is none for this function!

We can also see that this function has no vertical asymptotes since it is continuous for all real $x$.

Sso now for any horizontal asymptotes. We find:

\(\displaystyle \lim_{x->\infty}e^{-2x^3+3x^2+1}=0\)

\(\displaystyle \lim_{x->-\infty}e^{-2x^3+3x^2+1}=\infty\)

And this tell us that when $x$ goes to \(\displaystyle \infty\) then $y$ will go to zero, hence we have the horizontal asymptote given by $y=0$.

Now I leave it to you to draw the graph!

Thanks MarkFL for improving my post!

Regards,

\(\displaystyle |\pi\rangle\)

I wanted to take some time and show how to sketch the graph of a function! If you see anything that is wrong, please PM me and I will correct it! I hope you enjoy, understand and learn!

1. An exponential function

Draw the graph of \(\displaystyle y=e^{-2x^3+3x^2+1}\)

Okay, the first thing I notice is that the graph will never cut the $x$-axis because for any real $x$:

\(\displaystyle e^{-2x^3+3x^2+1}\neq0\)

If we look at y intercept ( Where the function cut the \(\displaystyle y\)-axis that means \(\displaystyle x=0\) so we get

\(\displaystyle e^{-2*0^3+3*0^2+1}=e\)

Okay, next we want to differentiate with respect to $x$ because we want to look at the slope.

So how do we differentiate that function?

Using the chain rule, we find:

\(\displaystyle \frac{dy}{dx}=e^{-2x^3+3x^2+1}\frac{d}{dx}(-2x^3+3x^2+1)=e^{-2x^3+3x^2+1}(-6x^2+6x)=6x(1-x)e^{-2x^3+3x^2+1}\)

If we want to find the critical values, we have to equate the derivative to zero (this is where the slope is zero)

Because \(\displaystyle e^{-2x^3+3x^2+1}\neq0\) this leaves us with:

\(\displaystyle 6x(1-x)=0\)

If we solve that equation we get \(\displaystyle x=1\) and \(\displaystyle x=0\) and it is at those $x$-values we will have extrema since they are roots of odd multiplicity and we therefore know the sign of the derivative will change across these critical values.

Let's make a schedule to analyze the intervals of increasing/decreasing behavior, and we see:

So, we may conclude that the given function is:

decreasing on \(\displaystyle (-\infty,0)\)

increasing on \(\displaystyle (0,1)\)

decreasing on \(\displaystyle (1,\infty)\)

Next, let's find the asymptotes.

Note: Only rational functions have oblique or slant asymptotes, so there is none for this function!

We can also see that this function has no vertical asymptotes since it is continuous for all real $x$.

Sso now for any horizontal asymptotes. We find:

\(\displaystyle \lim_{x->\infty}e^{-2x^3+3x^2+1}=0\)

\(\displaystyle \lim_{x->-\infty}e^{-2x^3+3x^2+1}=\infty\)

And this tell us that when $x$ goes to \(\displaystyle \infty\) then $y$ will go to zero, hence we have the horizontal asymptote given by $y=0$.

Now I leave it to you to draw the graph!

Thanks MarkFL for improving my post!

Regards,

\(\displaystyle |\pi\rangle\)

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