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C3

The question is :

f : x → ae x + a, x ∈ . Given that a is a positive constant, a sketch the graph of y = f(x), showing the coordinates of any points of intersection with the coordinate axes and the equations of any asymptotes.

I got the graph and its intersection correct but can someone explain why the equation of the asymptote is y=a.
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simon0
13
Noticed for the graph of the exponential where as you go further left (negative x direction), e^x gets closer to 0?

Note that for:

f(x)=ex, where x∈R, \displaystyle f(x) = e^{x}, \, \text{where}\, x \in \mathbb{R},

then:

lim⁡x←−∞f(x)=0. \displaystyle \lim_{x \leftarrow -\infty} f(x) = 0.

So for:

f(x)=aex+a (where x,a∈R and a is constant), lim⁡x←−∞f(x)=0+a=a. \displaystyle f(x) = ae^{x} + a\, \text{(where } x, a \in \mathbb{R} \text{ and a is constant)}, \, \lim_{x \leftarrow -\infty} f(x) = 0 + a = a.
(edited 6 years ago)
Original post by simon0
Noticed for the graph of the exponential where as you go further left (negative x direction), e^x gets closer to 0?

Note that for:

f(x)=ex, where x∈R, \displaystyle f(x) = e^{x}, \, \text{where}\, x \in \mathbb{R},

then:

lim⁡x←−∞f(x)=0. \displaystyle \lim_{x \leftarrow -\infty} f(x) = 0.

So for:

f(x)=aex+a (where x,a∈R), lim⁡x←−∞f(x)=0+a=a. \displaystyle f(x) = ae^{x} + a\, \text{(where } x, a \in \mathbb{R} \text{)}, \, \lim_{x \leftarrow -\infty} f(x) = 0 + a = a.


Damn I was so close, I get it now thank you

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