Respuesta :

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:

[tex]y = e ^ x[/tex]

Doing [tex]x = 0[/tex]we have:

[tex]y = e ^ 0[/tex]

Every number raised to zero power is 1.

So, we have:

[tex]y = 1[/tex]

We have the point (0,1)

If we try option D:

[tex]y = e ^ x + 1[/tex]

Doing x = 0 we have:

[tex]y = e ^ 0 + 1\\y = 1 + 1\\y = 2[/tex]

We have the point (0,2)

So, we note that the correct option is option C

ANswer:

Option C

Answer: Third option.

Step-by-step explanation:

Observe that the graph tends to zero when [tex]x<0[/tex] and "y" tends to infinite when [tex]x>0[/tex]

The function [tex]y=lnx[/tex] 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 [tex]y=1[/tex], while the function [tex]y=e^x+1[/tex] cuts the y-axis at [tex]y=2[/tex], therefore the equation of the last option does not represent that graph.

Finally, the equation that represents the graph is:

[tex]y=e^x[/tex]

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