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Which Function?

Alignments to Content Standards: F-IF.C.8 F-IF.C.8.a

Which of the following could be the function of a real variable $x$ whose graph is shown below? Explain.

 $f_1(x)=(x+12)^2+4$ $f_5(x)=-4(x+2)(x+3)$ $f_2(x)=-(x-2)^2-1$ $f_6(x)=(x+4)(x-6)$ $f_3(x)=(x+18)^2-40$ $f_7(x)=(x-12)(-x+18)$ $f_4(x)=(x-12)^2-9$ $f_8(x)=(24-x)(40-x)$

IM Commentary

The task addresses knowledge related to interpreting forms of functions derived by factoring or completing the square. It requires students to pay special attention to the information provided by the way the equation is represented as well as the sign of the leading coefficient, which is not written out explicitly, and then to connect this information to the important features of the graph. Students who have had plenty of experience re-writing quadratic expressions will be able to determine the sign of the leading coefficient without actually writing the quadratic expressions in $ax^2+bx+c$ form.

Solution

All of these are expressions for quadratic functions. Since quadratic functions have graphs that are parabolas and the given graph appears to be a parabola, the given expression meet a minimum criteria for consideration.

The graph of $f_1(x)=(x+12)^2+4$ has a vertex of $(-12, 4)$ which is in the second quadrant, so it does not match the graph.

The graph of $f_2(x)=-(x-2)^2-1$ has maximum rather than a minimum value at $x=2$ since the leading coefficient is negative (in other words, the graph opens downward), so it does not match the graph.

The graph of $f_3(x)=(x+18)^2-40$ has a vertex of $(-18, -40)$ which is in the third quadrant, so it does not match the graph.

The graph of $f_4(x)=(x-12)^2-9$ has a vertex of $(12, -9)$ which is in the fourth quadrant, and the leading coefficient is positive (so the graph would open upward) so this could describe the function whose graph is given.

The graph of $f_5(x)=-4(x+2)(x+3)$ has $x$-intercepts of $(-2,0)$ and $(-3,0)$. Since the $x$-intercepts are both positive for the given graph, they do not match.

The graph of $f_6(x)=(x+4)(x-6)$ has $x$-intercepts of $(-4,0)$ and $(6,0)$. The $x$-intercepts are both positive for the give graph, so they do not match.

The graph of $f_7(x)=(x-12)(-x+18)$ has a leading coefficient that is negative and so has a maximum rather than a minimum value (at $x=15$) and thus cannot match the graph.

The graph of $f_8(x)=(x-24)(x-40)$ has $x$-intercepts of $(24,0)$ and $(40,0)$. Since the $x$-intercepts are both positive for the graph and the leading coefficient is positive (so the graph would open upward), this could possibly be the equation for this graph.

The functions $f_4$ and $f_8$ both have graphs of the approximate shape given, though we note that they certainly would not appear identical if plotted simultaneously. For example, the vertex of the graph of $f_4$ occurs at $x=12$ whereas that of $f_8$ occurs at $x=32$.

Robert Hawke says:

over 4 years

Cam - with this fix, I think the two solutions are a little more plausible, but I still think kids will have somewhat competing priorities when responding to this task. Really, you're asking them to look at essential features of the parabola, and how those essential features are reflected - or not reflected - in each expression. The issue is that something you (and I, and most math teachers) see as inessential are important to students. For example, if you look at the scale between the y-axis and the x-intercepts, those (with the recent edits) now line up better, but there's still the issue of the scale on the y-axis. If you narrow the task too much, you lose a lot of what makes it good (which it is, for all the reasons you listed above.) Perhaps this could be resolved simply by changing the framing of the question from "which of the following could be...?" to "which of the following could not be..." Framing this in the negative would take away students' opportunity to discover or creatively apply the strategy of eliminating the possibilities, but it would give you more of a window into the features they (rightly) might consider essential, such as the unstated scale of the axes, and would give students a reason to make these assumptions explicit. This could, in turn, lead to a more rich discussion about the various features of this parabola.

I can see two other potential changes, if you don't like the change proposed above:

1. Change the scale so that only one expression works, and make the scale work exactly.
2. Explicitly call out the features of the parabola you want students to focus on, such as the x-intercepts. For what it's worth, I don't love this change, as I think it takes away the charm of this task, but it would make the task more clear for those who would otherwise focus on less salient aspects of this graph.

Cam says:

over 4 years

Hi Robert,

Thanks for the feedback! Am I right in understanding that your principal objection is that the scale on the $x$ axis is different from the scale on the $y$ axis? I can't speak for the author of the task, but I was envisioning the point being made that frequently we look at graphs where the scales on the axes are different -- indeed, this is the principal view that students get when they attempt to graph the functions on their calculators! (On the other hand, the problem resolved by the recent changes where that even the scale on the x-axis alone was inconsistent, which was definitely bad).

I also agree that it would be best, or at least in keeping with the sprit of the original problem, if we didn't overspecify the scales, to as to keep the focus on isolating the qualitiative features of parabolas, rather than just numerically plotting them.

Robert Hawke says:

over 4 years

Thanks for your reply. This definitely helps clarify the difference between the scales on the x- and y- axes. The issue still at play, however, is the precision of the scale on the x-axis. Though f8 has intercepts at 24 and 40 and f4 has intercepts at roughly the same spacing from zero, the spacing isn't exactly the same, and thus this couldn't actually be a graph of one AND the other...it's close, yes, but the two intercepts can't simultaneously satisfy two different ratios, in terms of distance from the y-axis. So this could be a graph of f4 but not f8, or of f8 but not f4. But, as graphed, it definitely could not be a graph of f4 and f8.

One way around this is to measure the actual distance, get the ratio of x-intercepts what you want it to be, and create two functions with this exact ratio of roots. If you want to use (24-x)(40-x), these have a ratio of 40/24, or 5/3. So another concave-up parabola with these same roots could be created by f4 = (x-12)^2 - 9.

This is a little nitpicky, for sure, but this dodges the issue of kids getting used to imprecise scales.

Cam says:

over 4 years

Aha! I concur completely (except for it being excessively nitpicky). I've made the change you suggested. Thanks again.

Bill says:

over 4 years

There's a problem with this task. None of the solutions work if you pay attention to the scale on the x-axis. Achievement First high school teachers.

over 4 years

Fixed. Thanks.