How to sketch the graph of a function

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SUMMARY

This discussion focuses on sketching the graph of the exponential function $$y=e^{-2x^3+3x^2+1}$$. Key points include that the graph does not intersect the x-axis, with a y-intercept at $$y=e$$. The first derivative, $$\frac{dy}{dx}=6x(1-x)e^{-2x^3+3x^2+1}$$, reveals critical points at $$x=0$$ and $$x=1$$, indicating intervals of increasing and decreasing behavior. The function has no vertical or oblique asymptotes, with a horizontal asymptote at $$y=0$$ as $$x$$ approaches infinity.

PREREQUISITES
  • Understanding of exponential functions
  • Familiarity with differentiation and the chain rule
  • Knowledge of critical points and their significance
  • Concept of asymptotes in functions
NEXT STEPS
  • Learn about the second derivative test for concavity
  • Study the properties of exponential functions in detail
  • Explore techniques for finding asymptotes in various types of functions
  • Practice sketching graphs of functions with different behaviors
USEFUL FOR

Students and educators in mathematics, particularly those studying calculus and graphing techniques for functions.

Petrus
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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 $$y=e^{-2x^3+3x^2+1}$$

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

$$e^{-2x^3+3x^2+1}\neq0$$

If we look at y intercept ( Where the function cut the $$y$$-axis that means $$x=0$$ so we get
$$e^{-2*0^3+3*0^2+1}=e$$

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:

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

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

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

$$6x(1-x)=0$$

If we solve that equation we get $$x=1$$ and $$x=0$$ 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:
2rhnqld.png


So, we may conclude that the given function is:

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

increasing on $$(0,1)$$

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

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:

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

$$\lim_{x->-\infty}e^{-2x^3+3x^2+1}=\infty$$

And this tell us that when $x$ goes to $$\infty$$ 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,
$$|\pi\rangle$$
 
Last edited:
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Re: How to draw a function

It might also be useful if you included the second derivative and concavity information.
 
Something else that can be useful for sketching the graph of a function is to find the $y$-intercept, if it has one. This gives you another point on the curve. :D
 
Hello,
I see have missed some stuff, I will make a second differentiate test ( I did take this problem from my book that was not interested on second differentiate test and forgot about that!) and about y intercept is something I totally forgot which is easy to se it will be $$y=e$$ Thanks for you all taking your time and helping me improve my post!

Regards,
$$|\pi\rangle$$
 

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