Which of these equations match the graph? Attachment 30395 It can be 1,2 or all of the options, it can also be none. I think that I and III work for the graph, but I might be wrong. Thanks!

Printable View

- Mar 12th 2014, 06:12 PMcanyouhelpExponential equation that matches graph?
Which of these equations match the graph? Attachment 30395 It can be 1,2 or all of the options, it can also be none. I think that I and III work for the graph, but I might be wrong. Thanks!

- Mar 12th 2014, 09:15 PMSlipEternalRe: Exponential equation that matches graph?
Why do you think that I and III work? Give some reasoning. It looks (estimating) that the limit as $x \to \infty$ is $-5$. Since the exponential function either approaches $0$ or $\infty$ as $x \to \infty$, it seems safe to assume $A \approx -5$. So, all three satisfy that. Next, if I is true, then $f(x) \approx -5 + Ba^x$. At $x=0, f(x) = -10$. So, we find $-10 = -5+Ba^0 = -5+B$. This means $B = -5$. However, according to I, $B > 0$, so it cannot be I.

Show reasoning for II and III. - Mar 14th 2014, 09:07 PMcanyouhelpRe: Exponential equation that matches graph?
- Mar 14th 2014, 09:15 PMromsekRe: Exponential equation that matches graph?
what you need to do is relax and use your brain more.

What can you see about that function? It goes to -infinity as x goes to negative infinity, it goes to -10 as x goes to infinity, f(0) is little less than -10.

Now plug -infinity, 0, and infinity into each of the functional forms for I, II, and III and see which one matches this.

Slow and methodical is the way to do mathematics unless you're Euler or Gauss, and you're not. - Mar 17th 2014, 11:22 AMSlipEternalRe: Exponential equation that matches graph?
I am going with the assumption that $\lim_{x \to \infty} f(x) \appox -5$ (I looked at the general trend, and I made that assumption. It doesn't matter if I am off by a little bit, as I am looking at trends). This means that we assume $A = -5$ for each of I, II, and III.

Next, I looked at the graph, and found that $f(0) \approx -10$ (looking at the graph again, I think it is closer to $f(0) \approx -12$, so I will use that value from now on). Then I plugged that into the equation. So, for II and III, I would recommend using that same value. So if II is correct, then $f(0) = A+Ba^{-x} \approx -5+Ba^{-0} = -5+B \approx -12$. Then $B \approx -7$ That does not rule out II. So, we need to check if $a>1$. To do that, I need to check another value for $x$. I approximate again. $f(1) \approx -10$. So, plugging in the value I just found for $B$:

$f(2) = A+Ba^{-1} \approx -5 -7a^{-1} \approx -10$

So, $a^{-1} \approx \dfrac{5}{7}$

Then taking reciprocals, $a \approx \dfrac{7}{5} = 1.4>1$. So, you cannot discount II as a possibility (it matches the trends present in the graph).

So, how about you show your work for evaluating the function given for III (and trying to see if you find any contradictions to the trends you see).