Question

Which equation is represented by the graph below?

Answer 1

Answer: Third option.

Step-by-step explanation:

Observe that the graph tends to zero when $x$ and "y" tends to infinite when $x>0$

The function $y=lnx$ is not defined for values of "x" less than zero, therefore the equations of the options 1 and 2 do not represent the graph attached.

Observe that the function cuts the y-axis at $y=1$, while the function $y=e^x+1$ cuts the y-axis at $y=2$, therefore the equation of the last option does not represent that graph.

Finally, the equation that represents the graph is:

$y=e^x$

Answer 2

For this case we discard options A and B, since by definition, the graph does not correspond to the function of the Neperian logarithm.

We tested option C:

$y = e ^ x$

Doing $x = 0$we have:

$y = e ^ 0$

Every number raised to zero power is 1.

So, we have:

$y = 1$

We have the point (0,1)

If we try option D:

$y = e ^ x + 1$

Doing x = 0 we have:

$y = e ^ 0 + 1\\y = 1 + 1\\y = 2$

We have the point (0,2)

So, we note that the correct option is option C

ANswer:

Option C