Instructions: Sketch the graphs of the following functions....

Instructions: Sketch the graphs of the following functions. Write neat solutions. Make sure each solution has the following information (if applicable): • Domain of $f(x)$. • x-intercepts and y-intercepts. If x-intercepts are hard to compute, then ignore them. • Vertical asymptotes. • Horizontal asymptotes. • Intervals where $f$ is increasing and decreasing. • Local minima and local maxima. • Intervals where $f$ is concave up and concave down. • Inflection points. • Use the above information to sketch the graph of $f$. Here is a suggested sequence to do this: -Begin by first marking the x and y intercepts. -Draw the vertical and horizontal asymptotes. -Investigate how the function approaches the vertical asymptotes (this helps you start the sketch). -Investigate where the function is increasing and decreasing and sketch a rough picture. -Investigate the concavity and adjust your sketch. -Make sure to label all of the local extrema and inflection points. $f(x) = frac{1}{x^2 - 4}$

Transcript

00:01 So in this question, we are given the following function, f of x equals 1 over x squared minus 4.

00:08 And we just want to sketch the graph and sort of give the details, the sort of analytic details.

00:14 Like where the asymptotes, where is an increase and decreasing, things like that.

00:19 And so first of all, the way that i do this is to analyze this thing algebraically first.

00:26 So we can factor the denominator into x plus 2 and x.

00:32 Minus 2 and that'll tell us a few things.

00:37 So first of all, what's the domain of this function? i'll write all these characteristics over in red.

00:46 So the domain are all the x -tiles you're allowed to plug in.

00:51 And basically when you have a fraction, like a rational function, right, when you need to do a division, you're just not allowed to divide by zero.

00:57 You're just need to figure out where might i divide by zero.

01:01 And i can see that just from the two factors, right? it's just when x plus 2 equal 0 or x minus 2 equals 0.

01:06 And so basically what that tells us is that x better not be equal to 2 or negative 2.

01:12 Right? at those two input values, we would give the division by 0 and that's bad.

01:17 So in interval notation, our domain here is negative infinity up until negative 2.

01:24 Union, right? that just means we have to leave out the value negative 2, but we can pick back up and go from negative 2 to positive 2.

01:31 We leave out positive 2 and go off to infinity.

01:34 So if there's three different.

01:36 That this function is defined on.

01:38 So there's our domain.

01:42 Let's see.

01:43 That also gives us an idea of the asymptotes.

01:45 So the vertical asymptote, which i'll know in va, happens where i have these divisions by zero.

01:51 And that's exactly x equals negative 2 and x equals 2.

01:54 So the vertical asymptotes happen at these lines, the vertical line x equals negative 2, as well as the vertical line x equals positive 2.

02:03 And as far as horizontal asymptotes go, right, we can look at the degrees of the polynomials in this rational function.

02:11 Now the numerator, it's just a constant function, so that's like a degree zero polynomial.

02:17 I'll write that here.

02:18 So it's like degree zero.

02:20 The denominator is an x squared.

02:22 That's degree two.

02:23 And just remember that when the degree of the denominator is larger than the degree of the numerator, that means we have a horizontal asymptote at y equals zero, sort of automatically.

02:35 So we have one horizontal asymptote at, y equals 0.

02:39 And so far, if i sketch the graph here in like the bottom right corner, what we have is i'll draw the asymptotes in red.

02:54 So we have a horizontal asymptote at y equals 0, and then we have these two vertical asymptotes at negative 2 and 2.

03:04 Now let's look at the x intercepts of this.

03:08 Where could this possibly equal 0? well, a fraction only 0 when the numerator equals 0 and the numerator is just a constant 1 right so there just are no x intercepts there are none right so you can set 1 over x squared minus 4 equal to 0 you try to solve algebraically there are no solutions to that right because you get 1 is supposed to 0 which is just impossible so there are no x intercepts what about the y intercepts well that's just plugging 0 right that's f of 0 so it's 1 over 0 minus 4 that's negative 1 4 right a negative 1 4 right a negative 0 so let me write that y intercept in blue.

03:48 I'll draw that as a blue dot here.

03:51 So it's at negative 0 .25.

03:55 All right.

03:56 And from here, we know that if i sketch the graph of this, the function is supposed to get close to the asymptotes.

04:04 And i just need to figure out, like, is it above the x -axis or below the x -axis? so let's look at this intercept.

04:09 If i plug in, like, positive 1 into my function, we get 1 over negative 3.

04:15 Which is a negative number.

04:18 I plugged negative one.

04:20 Again, that's just one over negative three.

04:24 Basically, i get something that's below the x -axis, and it's supposed to follow these asymptotes, right? so it's supposed to get close to these vertical asymptotes.

04:32 So i get a little upside -down parabola, like a downward facing parabola shape, in between my vertical asymptotes.

04:38 And that's just by thinking like, okay, you know, the behavior of this function need to get close to the asymptotes.

04:45 I can't go above the x -axis because there's a horizontal asymptote there.

04:50 And so i can just check to see if the function values are negative, which they are.

04:54 So i can graph that there.

04:55 You can also do some further calculus, right, like finding first and second derivatives to understand increasing or concavity.

05:03 But for this question, you actually just really need to look at the graph just by doing this process here.

05:09 Okay, so what about to the left of x equals negative 2.

05:13 So if i go over to x equals negative 3 and i plug that value in, let's just do that little calculation up here.

05:21 F of negative 3, that's 1 over negative 3 squared minus 4...

Question

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