(Pre-Calculus) How would you graph $y = (x^2 − 1)(x − 2)^2$ by hand?

Question

I've been doing some summer practice assignments for my upcoming calculus class, and I have been tasked with graphing $y = (x^2 − 1)(x − 2)^2$ by hand.

At first I started making a table of values, but I quickly realized both $x = 1$ and $x =2$ have $y$ values of $0,$ with a relative maximum in between. I am struggling to figure out how to find relative min/max without the use of my calculator and without the use of differential calculus (thanks google) as I have not learned this yet. If you were in my shoes, how would you go about solving this? (Without the use of a calculator).

Answer 1

Finding the zeros is a good first step and that's really easy in your case: in addition to the ones you found there is also $x = -1$. You should also note that $x = 2$ is a double zero. Since the degree of the polynomial is 4, that's all the roots.

Finding the behavior at very large (positive and negative) values of x is the next step. In your case, the function goes to plus infinity at either end, because of the squares.

Finding the y-intercept might be helpful as well: that's the point where the graph crosses the y-axis. Just set $x=0$ and calculate the value of $y$ there.

Finding maxima and minima would be the next step but since you have not had calculus yet, you can't use derivatives. But you can see that your function is negative between $x=-1$ and $x=1$ because the first factor is negative there (the second factor is always positive - or zero). So you know that the function is positive when $x<-1$, negative when $ -1 < x < 1$ and positive when $x > 1$, except at $x=2$ where it has a double zero. At the double zero, the function just kisses the x-axis - that's true for a zero of multiplicity higher than 1 and the higher the multiplicity the flatter the curve is around that zero.

That should be enough to let you sketch the function.

Answer 2

You did well! Hopefully, you can spot the roots from the factors. $y$ will be zero when either $x=2$ or $x=\pm1$, so those are our x-intercepts. For bonus points, recognize that the exponent on the $(x-2)$ term is $2$. Since that's even, it means that the graph will just "kiss" the x-axis instead of passing through it.

Maybe for fun, you can work out the y-values when x is $-2, 0, 1.5, 3$ just to have a sense of how quickly the graph grows around there, but that will be a fine graph for now. (Note that these are easy to calculate since it's factored -- don't multiply out that polynomial!) When you start calculus, you'll get a better idea of how to find the turning points, which would have helped you figure out the exact minimum point of the graph and where the maximum in [1,2] is. Enjoy that!

Answer 3

You have $x$ intercepts at $x=-1,1,2$ with $x=2$ being a double root.

Your $y$ intercept is $(0,-4)$

Your function is continuous and tends to infinity at both ends.

That gives you an outline and some extra points should help you to have a better understanding of the graph.

Answer 4

Completely factorise into linear factors $$(x-1)(x+1)(x-2)(x-2).$$ It is easy to see that the roots are $-1,\,1$ and $2.$ Now, consider the signs of each of the factors in the intervals $(-\infty,-1],\,[-1,1],\,[1,2]$ and $[2,+\infty)$ and then the product of the signs. This will give you a rough idea of how the function behaves on the real axis. You may also find the intercept on the $y$-axis (by setting $x=0$ for more definiteness.)