Graphs of Functions: Free Response

A product is $0$ exactly when one of its factors is $0$, so set each factor to $0$ in turn.

$x + 3 = 0 \Rightarrow x = -3, \qquad (x - 5)^2 = 0 \Rightarrow x = 5.$

The third factor never reaches $0$, and no solving is needed to see it. A square is never negative, so adding $4$ leaves a total that cannot fall below $4$:

$x^2 + 4 \ge 4 > 0 \quad \text{for every real } x.$

The only zeros are therefore $-3$ and $5$. The factor $x^2 + 4$ can be set aside for the rest of the question: with no root it never splits an interval, and being always positive it never changes a sign.

The zeros $-3$ and $5$ cut the line into three intervals. A linear factor changes sign only at its own root, and no root lies strictly inside any of these intervals, so every factor holds one fixed sign across each of them.

Two of the three factors are settled once and for all. The factor $(x - 5)^2$ is a square, so it is positive at every input except $5$, and $x^2 + 4$ is positive everywhere by part A. Only $(x + 3)$ ever turns negative.

For $x < -3$, that one factor is negative and the other two are positive:

$(-)(+)(+) = -, \qquad \text{so } f < 0 \text{ on } x < -3.$

For $-3 < x < 5$, and again for $x > 5$, the factor $(x + 3)$ has turned positive while the other two are unchanged:

$(+)(+)(+) = +, \qquad \text{so } f > 0 \text{ on } -3 < x < 5 \text{ and on } x > 5.$

Every verdict came from multiplying three signs. Nothing was read off a curve, and no input was substituted.

Compare the two factors that vanish, one zero at a time.

At $x = -3$ the responsible factor is $(x + 3)$, to the first power. It is negative for $x < -3$ and positive for $x > -3$, while the other two factors keep their signs across $-3$, so the product's sign is different on the two sides:

$(-)(+)(+) = - \quad \longrightarrow \quad (+)(+)(+) = +.$

A function that is negative just to the left of a zero and positive just to the right sits on opposite sides of the axis there, so its graph crosses.

At $x = 5$ the responsible factor is $(x - 5)^2$, an even power. A square is never negative, so

$(x - 5)^2 > 0 \quad \text{for every } x \ne 5,$

and this factor contributes a $+$ on both sides of $5$. The other factors also keep their signs across $5$, so the product is positive on both sides: the graph reaches the axis at $(5, 0)$ and returns to the side it came from. It touches rather than crosses.

Notice what the argument never used. At no point did it claim that a curve has to pass through zero in order to change sides. The sign on each side was computed from the factors, and the crossing is the conclusion drawn from those two computed signs, not the reason for them.

Evaluate $g$ at the two sampled inputs. Each time one factor of the product is $0$, so the whole product drops out and only the line's value survives.

$g(2) = 7 + c(0)(-3) = 7, \qquad g(5) = 13 + c(3)(0) = 13.$

Neither result mentions $c$ at all, so every choice of $c$ produces a function matching both rows.

Now compare $g$ with the line away from those inputs. Subtracting the line from $g$ leaves exactly the added product:

$g(x) - (2x + 3) = c(x - 2)(x - 5).$

Take any input other than $2$ and $5$. Then $x - 2 \ne 0$ and $x - 5 \ne 0$, and if $c \ne 0$ the right side is a product of three nonzero numbers. A product of nonzero numbers is never $0$, so

$g(x) \ne 2x + 3 \qquad \text{for every } x \ne 2, 5.$

Different nonzero constants $c$ give different functions by the same subtraction, so the two table rows are shared not by one function but by infinitely many.

Expand the product first, then collect like terms.

$g(x) = 2x + 3 + (x - 2)(x - 5) = 2x + 3 + x^2 - 7x + 10.$

Collecting gives a single quadratic rule:

$g(x) = x^2 - 5x + 13.$

Check it against the table before going on: $g(2) = 4 - 10 + 13 = 7$ and $g(5) = 25 - 25 + 13 = 13$, both correct. Now evaluate the two rules at $0$, an input the table never sampled:

$g(0) = 13, \qquad 2(0) + 3 = 3.$

The two rules agree exactly on the table and sit $10$ apart at an input the table never reached. A quadratic and a line are not remotely the same shape, and yet these two rows cannot tell them apart.

Separate two different claims that the classmate has run together with the word so.

The first claim is about lines: given two points with different first coordinates, exactly one line passes through both. That is correct, and here it really does single $2x + 3$ out.

The second claim is about $f$: that $f$ is a line. Nothing in the table says so. The table reports two values and stops, and part A built, for every constant $c$, a different function returning those same two values:

$g(x) = 2x + 3 + c(x - 2)(x - 5), \qquad g(2) = 7, \quad g(5) = 13.$

So the honest statement is that the two rows determine a unique line through the two points, and determine nothing whatever about $f$ at inputs other than $2$ and $5$. Uniqueness inside a family of functions is not uniqueness among all functions, and it is the second that would be needed to name $f$. This is why a plotted table is a starting point rather than evidence: the features of a graph have to be proved from a rule, and here no rule has been established.

A product is $0$ exactly when one of its factors is, so the model reports break-even at

$x = 0, \qquad x = 3, \qquad x = 10,$

all three inside the offered window. Those months cut the window into three stretches, and no factor has a root strictly inside any of them, so each factor holds one sign across each stretch and the sign of $P$ is the product of the three.

For $0 < x < 3$, the first factor is positive while the other two are still negative:

$(+)(-)(-) = +.$

For $3 < x < 10$, the middle factor has turned positive and the last has not:

$(+)(+)(-) = -.$

For $10 < x \le 12$, all three factors are positive, so the product is positive. Reading that back into the situation: the workshop is in profit for its first three months, runs at a loss from month $3$ to month $10$, and is back in profit for the remainder of the window.

Expand the product first, so the highest power is visible.

$P(x) = x(x^2 - 13x + 30) = x^3 - 13x^2 + 30x.$

Now take the highest power out of every term:

$P(x) = x^3\left(1 - \frac{13}{x} + \frac{30}{x^2}\right).$

Bound the bracket from below. The term $\frac{30}{x^2}$ is positive for every nonzero $x$, so dropping it can only make the bracket smaller:

$1 - \frac{13}{x} + \frac{30}{x^2} > 1 - \frac{13}{x}.$

Now pick the threshold. Once $x \ge 26$ we have $\frac{13}{x} \le \frac{13}{26} = \frac{1}{2}$, so the bracket is greater than $1 - \frac{1}{2}$, and multiplying by the positive factor $x^3$ gives

$P(x) > \frac{1}{2}x^3 \qquad \text{for every } x \ge 26.$

Name any bound you like; $x^3$ eventually exceeds twice it, and from that input onward $P$ exceeds the bound as well. No limit was taken anywhere. The whole argument is one inequality with a threshold written down beside it.

Part B produced a genuine inequality about the formula:

$P(x) > \frac{1}{2}x^3 \qquad \text{for } x \ge 26.$

Read it back into the situation. At twenty-six months the model already predicts more than $8788$ hundred dollars of profit in a single month, and its prediction keeps climbing past any figure you care to name. A workshop's monthly profit is capped by its floor space, its staff and the customers who exist, so a rule whose outputs have no ceiling cannot describe it indefinitely.

So the honest reading splits the work in two. The algebra establishes something about the formula: the values of $P$ pass every bound. It establishes nothing about the workshop, because the workshop was never claimed to obey $P$ outside $0 \le x \le 12$. The window is part of the model rather than a detail to be dropped once the formula is written down, and end behavior is precisely the calculation that shows why dropping it would be reckless.

Both definitions compare $f(-x)$ with $f(x)$ at every input of the domain, so both quietly require $-x$ to be in the domain whenever $x$ is. Check that requirement before anything else.

The denominator $x - 6$ is $0$ at $x = 6$, and nowhere else, so

$\text{domain of } f: \ x \ne 6.$

Now test the requirement at a single input. The number $-6$ is in the domain, since $-6 \ne 6$. Its negative is $6$, which is not:

$f(-6) = \frac{-216}{-12} = 18, \qquad f(6) \ \text{is undefined}.$

At the input $-6$, then, the expression $f(-x)$ has no value at all, so neither $f(-x) = f(x)$ nor $f(-x) = -f(x)$ can be written down there, never mind checked. The domain is not symmetric about $0$, and until it is, asking whether $f$ is even or odd is asking a question with no content. The point is worth stating carefully: the answer is not that $f$ is neither. It is that the test does not apply.

Find the excluded inputs from the denominator.

$x^2 - 36 = 0 \quad \Longrightarrow \quad x = 6 \ \text{ or } \ x = -6.$

This time the two are excluded together, so an input is allowed exactly when its negative is: the domain is symmetric about $0$ and the test applies. Substitute $-x$ into the whole rule, wrapping the argument in brackets before cubing and squaring:

$g(-x) = \frac{(-x)^3}{(-x)^2 - 36} = \frac{-x^3}{x^2 - 36}.$

The denominator came back unchanged because a square destroys the sign, while the numerator picked up a minus because a cube keeps it. Hence

$g(-x) = -\frac{x^3}{x^2 - 36} = -g(x),$

which is the definition of odd. The two rules differ by one factor in the denominator, and that single difference is what decides whether the question can be asked at all.

Start with the half of the conclusion that is right. Suppose $h$ is odd and $0$ is in its domain. Setting $x = 0$ in $h(-x) = -h(x)$ gives

$h(0) = -h(0), \qquad \text{so } 2h(0) = 0 \ \text{ and } \ h(0) = 0.$

An odd function whose domain contains $0$ therefore does pass through the origin, and $g$ from part B is one such function.

The conclusion overreaches by dropping the condition. Consider

$h(x) = \frac{x^2 + 1}{x}, \qquad x \ne 0.$

Its domain is every real number except $0$, which is symmetric about $0$, so the test applies, and

$h(-x) = \frac{(-x)^2 + 1}{-x} = -\frac{x^2 + 1}{x} = -h(x),$

so $h$ is odd. But $0$ is not an input at all, so the graph carries no point above or below it and cannot pass through the origin. The corrected statement is conditional: an odd function passes through the origin exactly when $0$ lies in its domain. The condition is doing real work, as this rule shows, so it is not a formality that can be trimmed away.

An input is excluded exactly when it makes the denominator $0$, so factor the denominator and set each factor to $0$. Two numbers multiplying to $12$ and adding to $7$ are $3$ and $4$:

$x^2 - 7x + 12 = (x - 3)(x - 4).$

A product is $0$ exactly when a factor is, so

$x - 3 = 0 \ \Rightarrow \ x = 3, \qquad x - 4 = 0 \ \Rightarrow \ x = 4.$

Both are rejected by the domain, and it matters that both are collected now. What the numerator turns out to be will change what each excluded input does to the picture, but nothing about the numerator can put either of them back into the domain.

Factor the numerator by taking out the common power.

$x^3 - 4x^2 = x^2(x - 4).$

Put it over the factored denominator from part A. The shared factor $(x - 4)$ cancels at every input where $f$ was defined in the first place, and only there:

$f(x) = \frac{x^2(x - 4)}{(x - 3)(x - 4)} = \frac{x^2}{x - 3}, \qquad x \ne 3, \ x \ne 4.$

So the graph of $f$ agrees with the curve $y = \dfrac{x^2}{x - 3}$ at every allowed input. The input $4$ is not allowed, so the point that curve occupies there is punched out. Its height comes from evaluating the simplified rule at that excluded input:

$\frac{4^2}{4 - 3} = \frac{16}{1} = 16.$

The missing point is $(4, 16)$. The other excluded input behaves quite differently: the factor $(x - 3)$ sits only in the denominator and survives the cancellation, so the simplified rule rejects $3$ as well and has no value there to reinstate. What the graph does near $3$ instead is a question for a later chapter; all that is needed here is that it is not one missing point.

The algebra is sound. The shared factor really does cancel, and at every input $f$ accepts, the two formulas return the same number.

The error lives in the word just. Cancelling changes what the FORMULA will accept: the simplified rule happily returns a value at $x = 4$, while $f$ does not, because $f$ was built from a denominator that rejects that input. A rule's domain is settled before any simplification, and simplifying never hands an input back.

$\text{at } x = 4: \quad \frac{x^2}{x - 3} \ \text{returns } 16, \qquad f \ \text{returns nothing.}$

So the correction is small but not cosmetic. Draw the simplified curve, then remove the single point it supplies at the cancelled input and mark that point as not belonging to the graph. This is what it means to say the domain is visible: the input that was excluded is invisible in the simplified formula and plain as a gap in the picture.