# Quartics

Investigate the graphs of y = [1 + (x - t)^2][1 + (x + t^)2] as the
parameter t varies.

Sketch graphs of

$$y = \left[1 + (x - t)^2\right]\left[1 + (x + t)^2\right]$$

for $t = -1/2$, $1/2$ and $2$. You will see that these graphs have 'different shapes'. Suppose the parameter $t$ varies, then the general shape of the graph varies continuously with $t$. Show that the graph always has a shape similar to the examples above and find the values of $t$ at which there are transitions from one shape to another.

For this question you may like to use a computer graph drawing application or a graphic calculator. For $t = -1/2$, $1/2$ and $2$, sketch graphs of $y = [1 + (x - t)^2][1 + (x + t)^2]$.

You can download the shareware program Graphmatica for free from here as NRICH is an approved distributor of this program. You can find more information about the program from http://www.graphmatica.com/

Then try other values of the parameter $t$. You will see that these graphs have 'different shapes'. Suppose the parameter $t$ varies, then the general shape of the graph varies continuously with $t$.

You can get this far without calculus but you'll probably need calculus to find the turning points and to show that the graph always has a shape similar to the examples above and to find the values of $t$ at which there are transitions from one shape to another.

Congratulations to Aleksander Twarowski from Gdynia Bilingual High School No 3, Poland for solving this Tough Nut. Well done!

Image

We can observe that for $t=2$ it has 3 stationary points (derivative is equal to 0), 2 minima and 1 maximum. It is an even function symmetrical about the $y$-axis. The function and its graph are the same for $t=-1/2$ as for $t=1/2$.

Let's expand this formula: $$\eqalign{ y&=[1+(x-t)^2][1+(x+t)^2] \cr &= [x^2 + (1+t^2) - 2tx][x^2 +(1+t^2)+2tx]\cr &= x^4 +2x^2(1+t^2)+(1+t^2)^2 - 4t^2x^2\cr &= x^4 +2(1-t^2)x^2 +(1+t^2)^2}$$ When we find its derivative with respect to $x$ we obtain $$y'=4x^3+4x(1-t^2)= 4x[x^2+(1-t)(1+t)].$$ Now we can investigate stationary points when $$4x[x^2+(1-t)(1+t)]=0.$$ It is important to observe that $[x^2+(1-t)(1+t)]$ can be factorized only when $(1-t^2)$ is negative or zero, that is $t^2\geq 1$. So there are three stationary points when $t$ belongs to interval from negative infinity to $-1$ and from $1$ to positive infinity. Hence, we can state that when $t$ belongs to $[-1,1]$ the graph looks like it does for $t=1/2$ above, and when $t$ belongs to $(-\infty,-1)$ or $(1,+\infty)$ the graphs have three stationary points and look like the one for $t=2$ above. Because these three intervals include all real values of $t$, the graphs cannot have other shapes.