In problems 13 32 graph the given function f by using...

Name: in problems 13 32 graph the given function f by using transformations to shift reflect shrink andor stretch the graphs of the basic functions fx 2x fx 3x fx 12x or fx 13x 13 fx 2x3 14 fx 3x1 23822
Uploaded: 2023-05-16T12:55:43-08:00
Duration: 7 min
Channel: Gregory Higby
Description: in problems 13 32 graph the given function f by using transformations to shift reflect shrink andor stretch the graphs of the basic functions fx 2x fx 3x fx 12x or fx 13x 13 fx 2x3 14 fx 3x1 23822

In problems 13-32, graph the given function $f$ by using transformations to shift, reflect, shrink, and/or stretch the graphs of the basic functions $F(x) = 2^x$, $F(x) = 3^x$, $F(x) = (1/2)^x$, or $F(x) = (1/3)^x$. 13. $f(x) = 2^{x+3}$. 14. $f(x) = -3^{x+1}$. 15. $f(x) = -(1/2)^{x+3}$. 16. $f(x) = -2^{-x}$. 17. $f(x) = -3(1/2)^x$. 18. $f(x) = (1/2)^{x+2}$. 19. $f(x) = -3^x+2$. 20. $f(x) = 3^{x-2}-9$. 21. $f(x) = (1/2)^{2x+2}$. 22. $f(x) = 3(1/2)^x$. 23. $f(x) = (1/3)^x-4$. 24. $f(x) = 3(1/2)^{x-1}$. 25. $f(x) = -2^{x+3}-2$. 26. $f(x) = -2^{x+2}+3$. 27. $f(x) = 3(1/2)^{x-2}$. 28. $f(x) = (1/3)^{x+1}-2$. 29. $f(x) = -2^{x-2}-2$. 30. $f(x) = -(1/2)^{x-3}$. 31. $f(x) = 2^{x-3}$. 32. $f(x) = e^{x-1}-1$. Answers to Problem Set 6.1 1. $2/3$. 3. $1/3$. 5. $-3$. 7. $-1$. 17. $f(x) = -3(1/2)^x$ $F(x) = (1/2)^x$ $G_1(x) = -F(x) = -(1/2)^x$ $f(x) = G_1(x)+3 = -3(1/2)^x$ 19. $f(x) = -3^x+2$ $F(x) = 3^x$ $G_1(x) = -F(x) = -3^x$ $f(x) = G_1(x)+2 = -3^x+2$

Transcript

00:01 I think the first thing i would do is graph these two functions.

00:04 And let me just start with a graph of 3 to the x, because its exponential growth looks something like that.

00:10 So what happens if you have g of x, i'm doing this in red, as 3 to the negative x power, is it reflects over the y -axis.

00:20 And what's different about that when you talk about f of x, which is 3 to the negative x minus 2 and then plus 1.

00:33 And i'm just looking at my graphing calculator real quick, because on this graph, it's hitting at 0, 1.

00:44 But on the other graph, that's actually going to be, i kind of want to just write it right here, f of x is equal to 3 to the negative x minus 2, and then minus 1 is taking this point and actually just shifts it a little up 2, up 1.

01:01 So it's going to hit the ordered pair negative 2, 2.

01:06 It doesn't look much different, but it is very different.

01:11 And that's a graph of f of x.

01:16 So to answer those questions then, i kind of just said it.

01:20 The graph of g to the graph of 1, it's just two shifts, left 2 and, wait, is it minus? i thought it was plus.

01:36 Yeah, i just calculated it down wrong.

01:43 So then the second question, write down the sets that represent the domain and range of f.

01:49 Ok, so the domain is all reals.

01:52 And i'm a big fan of integral notation, so i write it like this.

01:56 And then range is all values for y that are bigger than 1.

02:02 May i write that off to the side in case you've never seen this notation before.

02:07 The domain is all reals, and y is greater than 1.

02:11 And the reason why is we have that shifting up 1, and our horizontal asymptote is at y equals 1.

02:19 And those are your answers for f of x, not g of x.

02:23 So then what they want you to do next is take that equation, 3 to the negative x minus 2 plus 1, and equal it to 243, which isn't too bad, because you can subtract 1 over those 244.

02:43 I knew it was going to be 243, which i believe is 3 to the fifth power.

02:50 And that way, the exponents have to be the same now, because the bases are the same.

03:00 And you can add 2 over, but then x would have to equal negative 7.

03:06 And if you type that into your calculator, you can actually look at your table and see that x is negative 7.

03:14 You get 244.

03:17 Ok, i accidentally did f already with this graph over here.

03:27 The only thing i didn't do is draw the horizontal asymptote.

03:30 And i haven't given you the x, the y -intercept either.

03:36 And what you'd have to do is plug in 0 for x.

03:39 And so the y -intercepts, let me just write it down here.

03:45 Y -intercept would be the ordered pair f of 0, which would equate to 3 to the negative second power plus 1, which would equal 3 squared is 9.

03:57 So it's 1 over 9 plus 1, which would be 10 ninths.