Let's approach this by understanding how an exponential works, to begin.
Any exponential in the form: [tex]f(x) = 3^{x}[/tex] will cut the y-axis once and once only. This is because the exponential graph is a monotonic increasing graph for all real values of x.
We can prove this by taking the derivative. When we take the derivative of a function, we are finding the slope or gradient of the function at an arbitrary point on the graph.
[tex]\text{Differentiating the function:} f'(x) = \frac{d(3^{x})}{dx}[/tex]
[tex]f(x) = e^{ln(3^{x})}[/tex]
[tex]f(x) = e^{xln3}[/tex]
[tex]f'(x) = ln(3) \cdot e^{xln3}[/tex]
Thus, we can conclude for any value of x, the gradient will always be increasing. Now, by plotting some critical values, we can find that the graph hits the y-axis once and once only at the value y = 1.
At this point, we've sort of have an idea as to what the graph will do, but the most critical points are the endpoints, when x approaches infinity and negative infinity. In order to understand what they do at those points, instead of plotting those points, let's use a limit.
[tex]\lim_{x \to \pm \infty}(3^{x})[/tex]
Limits are a crucial tool in understanding what a graph does as the terms get arbitrarily huge and infinitesimally small. We can see that as x approaches infinity, we approach some really big number faster than a quadratic, linear line, and any function other than factorials.
However, let's approach the negative side. If we were to plug in negative infinity into the function, we get:
[tex]f(-\infty) = 3^{-\infty}[/tex]
[tex]f(-\infty) = \frac{1}{3^{\infty}}[/tex]
Now, we know that for a really big number in the denominator, the function will shrink smaller and smaller to a value, 0. Because the numerator is positive, we can see that the graph will NEVER hit below 0. So, we can conclude that the graph has a limit at x ≠ 0.
Since we know the endpoints, the gradient at any value, and the y-intercept, we can graph the function. Refer to the other answer for the correct graph of 3ˣ.
