Graphs of Functions: Free Response

Every point on the graph of $q$ has the form $(x, q(x))$: the input is the first coordinate, and the output is the second. Reading straight from the list,

$q(-4) = 6 \;\Rightarrow\; (-4, 6), \qquad q(0) = -5 \;\Rightarrow\; (0, -5), \qquad q(5) = 15 \;\Rightarrow\; (5, 15).$

No computation is needed here, only writing each given fact in the shape of a point.

Solving $q(x) = 2$ means finding every input whose paired output is $2$, so scan the whole list rather than stopping at the first match.

$q(-4)=6,\ q(-2)=2,\ q(0)=-5,\ q(1)=2,\ q(3)=7,\ q(5)=15.$

Two entries give the output $2$: the input $-2$ and the input $1$. So $q(x) = 2$ has two solutions among the six listed values, not one.

A function's rule assigns each input to exactly one output; that is the definition. So the same input $x = 0$ cannot appear on two different rows of any table for $q$ with two different outputs, whichever value $q(0)$ is, it is the only one.

$q(0) = -5 \quad\Rightarrow\quad \text{y-intercept } (0, -5).$

Nothing in that rule stops two DIFFERENT inputs from sharing one output. Different rows may still land on the same output, with no rule broken, which is exactly the kind of repeat part B asked about.

Sweep the picture from left to right. The curve begins at $x = -4$, but that end is drawn as an open circle, so $x = -4$ is NOT included. It ends at $x = 3$, drawn as a filled dot, so $x = 3$ IS included. Every input between those two ends is covered, giving

$\text{domain: } -4 < x \le 3.$

Sweep the picture from bottom to top. The lowest point the curve reaches is a valley, and the highest is a peak; both of those extreme heights sit in the middle of the curve, away from either end. Reading their heights off the vertical axis gives

$\text{range: } -2 \le y \le 4.$

The open endpoint's height, read directly off the picture, is $-1$. Comparing it to the range found in part B,

$-2 \le -1 \le 4,$

shows $-1$ sits strictly inside that interval, nowhere near either extreme. Because the curve is unbroken, it must pass through every height between its lowest and highest point somewhere along its interior, so the height $-1$ is reached again at an included input. Excluding the one input that is open therefore removes nothing from the range.

Both readings come straight from the recorded list. Two seconds after release,

$H(2) = 18 \text{ m}.$

The ball lands the instant its height reaches $0$, and the list shows that happening at

$t = 5 \text{ s}.$

Scan every recorded height, not only the first one that matches.

$H(0) = 10, \quad H(1)=16,\quad H(2)=18,\quad H(3)=16,\quad H(4)=10,\quad H(5)=0.$

The height $10$ meters appears twice: once at $t=0$ and again at $t=4$. The ball passes through the same height once on the way up and once on the way down.

Time $t$ in this context counts seconds after the ball leaves the thrower's hand, so $t=0$ marks that release, and

$t < 0 \quad\Longrightarrow\quad \text{before the ball was ever in flight.}$

A numerical pattern of heights could still be evaluated for negative $t$ on paper, but no such $t$ names an actual second of THIS flight, so the domain of $H$ as a real-world function stops at $t=0$: it is $0 \le t \le 5$, not a wider set.

A turning point is where the curve stops rising and starts falling, or the reverse. Reading the two marked points off the grid,

$\text{first turning point} = (-2, 3), \qquad \text{second turning point} = (1, -3).$

At $(-2,3)$ the curve arrives rising and leaves falling, so that point is a maximum. At $(1,-3)$ it arrives falling and leaves rising, so that point is a minimum.

Read the curve left to right. Before the first turning point it climbs, so $k$ is increasing on $x < -2$. Between the two turning points it falls, so $k$ is decreasing on $-2 < x < 1$. After the second turning point it climbs again, so $k$ is increasing on $x > 1$.

$\text{increasing: } x<-2 \text{ or } x>1, \qquad \text{decreasing: } -2<x<1.$

The curve has three monotonic branches, separated by the turning points at heights $3$ (the maximum) and $-3$ (the minimum): rising from far below up to $3$, falling from $3$ down to $-3$, then rising from $-3$ onward. A height $b$ that sits strictly between the minimum and the maximum is crossed exactly once by each of the three branches, since each branch is monotonic and spans past that height. Here $b=0$, and

$-3 < 0 < 3,$

so $0$ lies strictly between the two turning-point heights. Each of the three branches therefore crosses $y=0$ exactly once, giving three solutions in total.

The middle branch runs only from $-3$ up to $3$ and never leaves that band, so it can never reach a height outside it. The two outer branches are each unbounded in only one direction: the left branch heads to $-\infty$ but tops out at $3$, and the right branch heads to $+\infty$ but bottoms out at $-3$. So for $b > 3$, only the right branch (heading to $+\infty$) ever reaches it, giving one solution; for $b < -3$, only the left branch (heading to $-\infty$) ever reaches it, again one solution. Once $b$ moves back inside the band between the two turning-point heights, all three branches are back in play, and the count jumps, as part C found for $b=0$.

$k(x)=b \text{ has exactly one solution} \iff b<-3 \text{ or } b>3.$

One short function is enough to break a claim about EVERY function, so build one on purpose: pick two different inputs and deliberately give them the same output. For instance,

$m(-3) = 5, \qquad m(0) = 8, \qquad m(2) = 5.$

This is a perfectly good function: each input still has only one output. But look at the equation $m(x) = 5$. Both $x=-3$ and $x=2$ satisfy it, so

$m(x) = 5 \quad\Longrightarrow\quad x=-3 \text{ or } x=2,$

two solutions from one output. The claim, that solving always gives exactly one answer, fails on this example.

Evaluating $m$ at a single input, say $x=-3$,

$m(-3) = 5,$

gives one number and nothing else. A function is DEFINED so that each input is assigned exactly one output; if $m(-3)$ could equal two different numbers at once, $m$ would not be a function in the first place. So this failure mode cannot happen, regardless of how the rest of $m$'s values are chosen.

The claim silently assumed the two directions behave alike, but the definition of a function is one-directional: it promises

$\text{each input} \longrightarrow \text{exactly one output},$

and says nothing at all about how many inputs may point to the same output. So the corrected claim keeps the guarantee for evaluating, since that direction really is fixed by the definition, but must allow for several answers when solving, since nothing rules those out.