## anonymous one year ago HELP ME PLEASE Which of the following exponential functions goes through the points (1, 6) and (2, 12)? f(x) = 3(2)x f(x) = 2(3)x f(x) = 3(2)−x f(x) = 2(3)−x

1. ybarrap

Plug in x=1 and see if it equals 6 Plug in x=2 and see if it equals 12 Make sense?

2. anonymous

not really? can you guide me through it? (xD I dont want to be an answer-hogger/wanter.. I dont want the answer, just explanations :) ) @freckles

3. anonymous

Do you understand that coordinates are in the form (x,y)?

4. anonymous

yes. @BMF96

5. anonymous

f(x) = y

6. anonymous

What don't you understand?

7. jim_thompson5910

If $\Large f(x) = 7(3)^x$ (for example), then what is the value of f(x) when x = 2? In other words, what is f(2) equal to?

8. anonymous

f(2)=441?? #CONFUSED

9. jim_thompson5910

Replace each x with 2 $\Large f(x) = 7(3)^x$ $\Large f(2) = 7(3)^2$ $\Large f(2) = 7(9)$ $\Large f(2) = 63$ Do you see how I got f(2) to be equal to 63?

10. jim_thompson5910

btw you square first and then you multiply

11. anonymous

ok, I forgot that rule :) (like DUHR, Bella, get a grip)

12. jim_thompson5910

Since $$\Large f({\color{red}{2}}) = {\color{blue}{63}}$$ from my example, this means the point $$\Large ({\color{red}{x}},{\color{blue}{y}})=({\color{red}{2}},{\color{blue}{63}})$$ lies on the function curve of f(x)

13. anonymous

okayyy....

14. anonymous

$$f(x)=y= \begin{cases} 3(2)^x\\ 2(3)^x\\ 3(2)^{-x}\\ 2(3)^{-x} \end{cases}\qquad \qquad \begin{array}{llll} x&y \\\hline\\ {\color{brown}{ 1}}&3(2)^{\color{brown}{ 1}}\\ &2(3)^{\color{brown}{ 1}}\\ &3(2)^{-{\color{brown}{ 1}}}\\ &2(3)^{-{\color{brown}{ 1}}}\\ {\color{brown}{ 2}}&3(2)^{\color{brown}{ 2}}\\ &2(3)^{\color{brown}{ 2}}\\ &3(2)^{-{\color{brown}{ 2}}}\\ &2(3)^{-{\color{brown}{ 2}}} \end{array}$$

15. jim_thompson5910

so what ybarrap said at the top, you plug in each x coordinate into each function and see if you get the correct corresponding y coordinates Let's say we pick choice B at random $\Large f(x) = 2(3)^x$ The first point is (1,6). To test if this point lies on the function f(x) curve, we plug in x = 1 and see if y = 6 pops out $\Large f(x) = 2(3)^x$ $\Large f(1) = 2(3)^1$ $\Large f(1) = 2(3)$ $\Large f(1) = 6$ It does, so (1,6) is definitely on this curve. How about (2,12)? Let's check $\Large f(x) = 2(3)^x$ $\Large f(2) = 2(3)^2$ $\Large f(2) = 2(9)$ $\Large f(2) = 18$ Nope. The point (2,18) actually lies on this function curve and NOT (2,12). So we can rule out choice B.

16. jim_thompson5910

So what just happened was that I've proven that the function $$\Large f(x) = 2(3)^x$$ does NOT go through both points (1,6) and (2,12).

17. anonymous

okay, so its A, C, or D lol.... so lets try to rule out A...

18. anonymous

@jim_thompson5910

19. jim_thompson5910

what did you get so far in checking choice A?

20. anonymous

that A is, in fact NOT the answer!?

21. jim_thompson5910

if x = 1, then what is f(1) ?

22. anonymous

f

23. anonymous

Wait, so it IS A!!!!!

24. jim_thompson5910

$\Large f(x) = 3(2)^x$ $\Large f(1) = 3(2)^1$ $\Large f(1) = \underline{ \ \ \ \ \ \ \ } \text{ (fill in the blank)}$

25. anonymous

6

26. jim_thompson5910

27. anonymous

12

28. jim_thompson5910

Good. Choice A is definitely the answer. As practice, why not go through C and D and eliminate them. With choice C, if x = 1, then what is f(x) equal to?