# Tag Archives: slider

## Systems of lines

Here’s an interesting variation of a typical (MS) problem I found by following the Five Triangles ‘blog: http://fivetriangles.blogspot.com/2013/09/97-no-triangle.html .

(Note:  If you sign up on this or other ‘blogs, you can get lots of problems emailed to you every time they are added.)

INITIAL SOLUTION

I know this question can absolutely be solved without using technology, but when a colleague asked if it was appropriate to use technology here (my school is one-to-one with tablet laptops), I thought it would be cool to share with her the ease and power of Desmos.  You can enter the equations from the problem exactly as given (no need to solve for y), or you can set up a graph in advance for your students and email them a direct link to an already-started problem.

If you follow this link, you can see how I used a slider (a crazy-simple addition on Desmos) to help students discover the missing value of a.

FOLLOW-UP

I suggest in this case that playing with this problem graphically would grant insight for many students into the critical role (for this problem) of the intersection point of the two explicitly defined lines.  With or without technology support, you could then lead your students to determine the coordinates of that intersection point and thereby the value of a.

Keeping with my CAS theme, you could determine those coordinates using GeoGebra’s brand new CAS View:

Substituting the now known values of x and y into the last equation in the problem gives the desired value of a.

NOTE:  I could have done the sliders in GeoGebra, too, but I wanted to show off the ease of my two favorite (and free!) online math tools.

CONCLUSION

Thoughts?  What other ideas or problems could be enhanced by a properly balanced use of technology?

As an extension to this particular problem, I’m now wondering about the area of triangle formed for any value of a.  I haven’t played with it yet, but it looks potentially interesting.  I see both tech and non-tech ways to approach it.

## Extending graph control

This article takes my idea from yesterday’s post about using $g(x)=\sqrt \frac{\left | x \right |}{x}$ to control the appearance of a graph and extends it in two ways.

• Part I below uses Desmos to graph $y=(x+2)^3x^2(x-1)$ from the left and right simultaneously
• Part II was inspired by my Twitter colleague John Burk who asked if this control could be extended in a different direction.

Part I: Simultaneous Control

When graphing polynomials like $y=(x+2)^3x^2(x-1)$, I encourage my students to use both its local behavior (cubic root at $x=-2$, quadratic root at $x=0$, and linear root at $x=1$) and its end behavior (6th degree polynomial with a positive lead coefficient means $y\rightarrow +\infty$ as $x\rightarrow\pm\infty$). To start graphing, I suggest students plot points on the x-intercepts and then sketch arrows to indicate the end behavior.  In the past, this was something we did on paper, but couldn’t get technology to replicate it live–until this idea.

In class last week, I used a minor extension of yesterday’s idea to control a graph’s appearance from the left and right simultaneously.  Yesterday’s post suggested  multiplying  by $\sqrt \frac{\left | a-x \right |}{a-x}$ to show the graph of a function from the left for $x.  Creating a second graph multiplied by $\sqrt \frac{\left | x-b \right |}{x-b}$ gives a graph of your function from the right for $b.  The following images show the polynomial’s graph developing in a few stages.  You can access the Desmos file here.

First graph the end behavior (pull the a and b sliders in a bit to see just the ends of the graph) and plot points at the x-intercepts.

From here, you could graph left-to-right or right-to-left.  I’ll come in from the right to show the new right side controller. The root at $x=1$ is linear, so decreasing the b slider to just below 1 shows this.

Continuing from the right, the next root is a bounce at $x=0$, as shown by decreasing the b slider below 0.  Notice that this forces a relative minimum for some $0.

Just because it’s possible, I’ll now show the cubic intercept at $x=2$ by increasing the a slider above 2.

All that remains is to connect the two sides of the graph, creating one more relative minimum in $-2.

The same level of presentation control can be had for any function’s graph.

Part II: Vertical Control

I hadn’t thought to extend this any further until my colleague asked if a graph could be controlled up and down instead of left and right.  My guess is that the idea hadn’t occurred to me because I typically think about controlling a function through its domain.  Even so, a couple minor adjustments accomplished it.  Click here to see a vertical control of the graph of $y=x^3$ from above and below.

Enjoy.

## Controlling graphs and a free online calculator

When graphing functions with multiple local features, I often find myself wanting to explain a portion of the graph’s behavior independent of the rest of the graph.  When I started teaching a couple decades ago, the processor on my TI-81 was slow enough that I could actually watch the pixels light up sequentially.  I could see HOW the graph was formed.  Today, processors obviously are much faster.  I love the problem-solving power that has given my students and me, but I’ve sometimes missed being able to see function graphs as they develop.

Below, I describe the origins of the graph control idea, how the control works, and then provide examples of polynomials with multiple roots, rational functions with multiple intercepts and/or vertical asymptotes, polar functions, parametric collision modeling, and graphing derivatives of given curves.

BACKGROUND:  A colleague and I were planning a rational function unit after school last week wanting to be able to create graphs in pieces so that we could discuss the effect of each local feature.  In the past, we “rigged” calculator images by graphing the functions parametrically and controlling the input values of t.  Clunky and static, but it gave us useful still shots.  Nice enough, but we really wanted something dynamic.  Because we had the use of sliders on our TI-nSpire software, on Geogebra, and on the Desmos calculator, the solution we sought was closer than we suspected.

REALIZATION & WHY IT WORKS: Last week, we discovered that we could use $g(x)=\sqrt \frac{\left | x \right |}{x}$ to create what we wanted.  The argument of the root is 1 for $x<0$, making $g(x)=1$.  For $x>0$, the root’s argument is -1, making $g(x)=i$, a non-real number.  Our insight was that multiplying any function $y=f(x)$ by an appropriate version of g wouldn’t change the output of f if the input to g is positive, but would make the product ungraphable due to complex values if the input to g is negative.

If I make a slider for parameter a, then $g_2(x)=\sqrt \frac{\left | a-x \right |}{a-x}$ will have output 1 for all $x.  That means for any function $y=f(x)$ with real outputs only, $y=f(x)\cdot g_2(x)$ will have real outputs (and a real graph) for $x only.  Aha!  Using a slider and $g_2$ would allow me to control the appearance of my graph from left to right.

NOTE:  While it’s still developing, I’ve become a big fan of the free online Desmos calculator after a recent presentation at the Global Math Department (join our 45-60 minute online meetings every Tuesday at 9PM ET!).  I use Desmos for all of the following graphs in this post, but obviously any graphing software with slider capabilities would do.

EXAMPLE 1:  Graph $y=(x+2)^3x^2(x-1)$, a 6th degree polynomial whose end behavior is up for $\pm \infty$, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1.

Click here to access the Desmos graph that created the image above.  You can then manipulate the slider to watch the graph wiggle through, then bounce off, and finally pass through the x-axis.

EXAMPLE 2:  Graph $y=\frac{(x+1)^2}{(x+2)(x-1)^2}$, a 6th degree polynomial whose end behavior is up for $\pm \infty$, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1.

Click here to access the Desmos graph above and control the creation of the rational function’s graph using a slider.

EXAMPLE 3:  I believe students understand polar graphing better when they see curves like the  limacon $r=2+3cos(\theta )$ moving between its maximum and minimum circles.  Controlling the slider also allows users to see the values of $\theta$ at which the limacon crosses the pole. Here is the Desmos graph for the graph below.

EXAMPLE 4:  Object A leaves (2,3) and travels south at 0.29 units/second.  Object B leaves (-2,1) traveling east at 0.45 units/second.  The intersection of their paths is (2,1), but which object arrives there first?  Here is the live version.

OK, I know this is an overly simplistic example, but you’ll get the idea of how the controlling slider works on a parametrically-defined function.  The $latex \sqrt{\frac{\left | a-x \right |}{a-x}}$ term only needs to be on one of parametric equations.  Another benefit of the slider approach is the ease with which users can identify the value of t (or time) when each particle reaches the point of intersection or their axes intercepts.  Obviously those values could be algebraically determined in this problem, but that isn’t always true, and this graphical-numeric approach always gives an alternative to algebraic techniques when investigating parametric functions.

ASIDE 1–Notice the ease of the Desmos notation for parametric graphs.  Enter [r,s] where r is the x-component of the parametric function and s is the y-component.  To graph a point, leave r and s as constants.  Easy.

EXAMPLE 5:  When teaching calculus, I always ask my students to sketch graphs of the derivatives of functions given in graphical forms.  I always create these graphs one part at a time.  As an example, this graph shows $y=x^3+2x^2$ and allows you to get its derivative gradually using a slider.

ASIDE 2–It is also very easy to enter derivatives of functions in the Desmos calculator.  Type “d/dx” before the function name or definition, and the derivative is accomplished.  Desmos is not a CAS, so I’m sure the software is computing derivatives numerically.  No matter.  Derivatives are easy to define and use here.

I’m hoping you find this technology tip as useful as I do.

## Cubics and CAS

Here’s a question I posed to one of my precalculus classes for homework at the end of last week along with three solutions we developed.

Suppose the graph of a cubic function has an inflection point at (1,3) and passes through (0,-4).

1. Name one other point that MUST be on the curve, and
2. give TWO different cubic equations that would pass through the three points.

SOLUTION ALERT!  Don’t read any further if you want to solve this problem for yourself.

The first question relies on the fact that every cubic function has 180 degree rotational symmetry about its inflection point.  This is equivalent to saying that the graph of a cubic function is its own image when the function’s graph is reflected through its inflection point.

That means the third point is the image of (0,-4) when point-reflected through the inflection point (1,3), which is the point (2,10) as shown graphically below.

From here, my students came up with 2 different solutions to the second question and upon prodding, we created a third.

SOLUTION 1:  Virtually every student assumed $y=a\cdot x^3$ was the parent function of a cubic with unknown leading coefficient whose “center” (inflection point) had been slid to (1,3).  Plugging in the given (0,-4) to $(y-3)=a\cdot (x-1)^3$ gives $a=7$.  Here’s their graph.

SOLUTION 2:  Many students had difficulty coming up with another equation.  A few could sketch in other cubic graphs (curiously, all had positive lead coefficients) that contained the 3 points above, but didn’t know how to find equations.  That’s when Sara pointed out that if the generic expanded form of a cubic was $a\cdot x^3+b\cdot x^2 +c\cdot x+d$ , then any 4 ordered pairs with unique x-coordinates should define a unique cubic.  That is, if we picked any 4th point with x not 0, 1, or 2, then we should get an equation.  That this would create a 4×4 system of equations didn’t bother her at all.  She knew in theory how to solve such a thing, but she was thinking on a much higher plane.  Her CAS technology expeditiously did the grunt work, allowing her brain to keep moving.

A doubtful classmate called out, “OK.  Make it go through (100,100).”  Following is a CAS screen roughly duplicating Sara’s approach and a graph confirming the fit.  The equation was onerous, but with a quick copy-paste, Sara had moved from  idea to computation to ugly equation and graph in just a couple minutes.  The doubter was convinced and the “wow”s from throughout the room conveyed the respect for the power of a properly wielded CAS.

In particular, notice how the TI-Nspire CAS syntax in lines 1 and 3 keep the user’s focus on the type of equation being solved and eliminates the need to actually enter 4 separate equations.  It doesn’t always work, but it’s a particularly lovely piece of scaffolding when it does.

SOLUTION 3:  One of my goals in Algebra II and Precalculus courses is to teach my students that they don’t need to always accept problems as stated.  Sometimes they can change initial conditions to create a much cleaner work environment so long as they transform their “clean” solution back to the state of the initial conditions.

In this case, I asked what would happen if they translated the inflection point using $T_{-1,-3}$ to the origin, making the other given point (-1,-7).  Several immediately called the 3rd point to be (1,7) which “untranslating” — $T_{1,3}(1,7)=(2,10)$ — confirmed to be the earlier finding.

For cases where the cubic had another real root at $x=r$, then symmetry immediately made $x=-r$ another root, and a factored form of the equation becomes $y=a\cdot (x)(x-r)(x+r)$ for some value of a.  Plugging in (-1,-7) gives a in terms of r.

The last line slid the initially translated equation using $T_{1,3}$ to re-position the previous line according to the initial conditions.  While unasked for, notice how the CAS performed some polynomial division on the right-side expression.

I created a GeoGebra document with a slider for the root using the equation from the last line of the CAS image above.  The image below shows one possible position of the retranslated root.  If you want to play with a live version of this, you will need a free copy of GeoGebra to run it, but the file is here.

What’s nice here is how the problem became one of simple factors once the inflection point was translated to the origin.  Notice also that the CAS version of the equation concludes with $+7x-4$, the line containing the original three points.  This is nice for two reasons.  The numerator of the rational term is $-7x(x-2)(x-1)$ which zeros out the fraction at x=0, 1, or 2, putting the cubic exactly on the line $y=7x-4$ at those points.

The only r-values are in the denominator, so as $r\rightarrow\infty$, the rational term also becomes zero.  Graphically, you can see this happen as the cubic “unrolls” onto $y=7x-4$ as you drag $|x|\rightarrow\infty$.  Essentially, this shows both graphically and algebraically that $y=7x-4$ is the limiting degenerate curve the cubic function approaches as two of its transformed real roots grow without bound.