# Why does an exponential function eventually get bigger than a quadratic

I have seen the answer to this question and this one.

My $$77$$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $$3x3^x$$ vs. $$10x210 x^2$$ that he can just try different integer values of $$xx$$ until he finds one, e.g. $$x=6x=6$$. But, how can a $$77$$th grader understand that it will always be true, even $$1.0001x1.0001^x$$ will eventually by greater than $$1000x21000 x^2$$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L’Hopital’s rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $$11$$. Although for base between $$00$$ and $$11$$, it is still true that there exists some $$xx$$ where the exponential is larger than the quadratic, the phrase “eventually” makes it sound like there is some $$MM$$ where it is larger for all $$x>Mx>M$$. So, I don’t like the way the question is written.

If you have a quadratic polynomial $$f(x)f(x)$$ and an exponential function $$b(x)=bxb(x) = b^x$$ where $$b>1b>1$$, you can show that $$b(x)b(x)$$ surpasses the polynomial by showing that eventually the growth rate of $$b(x)b(x)$$ exceeds the polynomials growth rate.

Since the polynomial’s leading term has the biggest effect when $$xx$$ grows very big, that means that the other terms don’t matter, so let $$f(x)=ax2f(x) = ax^2$$ for an $$a>0a>0$$. Now, compute the ratio between $$f(x+1)f(x+1)$$ and $$f(x)f(x)$$: $$f(x+1)f(x)=a(x+1)2ax2=(x+1x)2. \frac{f(x+1)}{f(x)} = \frac{a(x+1)^2}{ax^2} = \left(\frac{x+1}{x}\right)^2 \, .$$
As you can see, as $$xx$$ grows very big, that ratio between $$x+1x+1$$ and $$xx$$ grows close to $$11$$, thus $$f(x+1)f(x+1)$$ barely increases from $$f(x)f(x)$$ because it is being multiplied by a number close to $$11$$. Now analyze the ratio between $$b(x+1)b(x+1)$$ and $$b(x)b(x)$$. By definition, the exponential function multiplies by its base $$bb$$ evey time you increase by $$11$$, so the ratio between $$b(x+1)b(x+1)$$ and $$b(x)b(x)$$ is always $$bb$$. However, we stated already that $$b>1b>1$$. We also found out that $$f(x)f(x)$$ ratio approaches $$11$$ as $$xx$$ gets really big. Thus, there is a point when $$b(x)b(x)$$ ratio exceeds $$f(x)f(x)$$ ratio, which means that $$b(x)b(x)$$ will start growing faster than $$f(x)f(x)$$ and will eventually outgrow $$f(x)f(x)$$.

Note that I just used terminology like approach and really big because a $$77$$th grader would not know of limits, so don’t nitpick that.

Secondary note: I said $$a>0a>0$$ and $$b>1b>1$$ because I assumed that both of the functions would be traveling upwards as you moved right along the $$xx$$-axis.

If you want a downward-facing parabola with $$a<0a<0$$ and a downward-facing exponential with $$0, then you can just note that the exponential will just tend towards $$00$$ when $$xx$$ gets very big, but the quadratic will eventually go below zero if it is facing downwards, thus showing that the exponential will eventually become greater then the quadratic.