EXAM TWO fresh from today

Graph → Equation

Today you called this "a flow chart" by your second problem — and you were right. Every vertical asymptote is a factor in the denominator. Every zero is a factor in the numerator — crosses through it, it's odd; bounces off it, it's even. The horizontal asymptote hands you the coefficient — and if that number's negative, the sign has to live somewhere: either out front, or absorbed into flipping a factor, like $(x-2)$ becoming $(2-x)$. And a horizontal asymptote that isn't zero is a promise: the degrees on top and bottom have to match — don't forget a bare $x$ counts as degree 1.

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Cold open no checklist yet

You Already Solved This One

This is the exact graph from this morning. Don't scroll to the flow chart — cover it in your head and build the equation from scratch on paper first. Then reveal to see it mapped out step by step.

Check your reconstruction ›

Denominator, from the verticals. $x=-3$ spikes the same direction on both sides — even power, $(x+3)^2$. $x=3$ flips sign across it — odd power, and since the horizontal asymptote is negative, that's exactly where the sign gets absorbed: $(3-x)^3$, not $(x-3)^3$.

Numerator, from the zeros. $x=-4$ crosses straight through — odd, $(x+4)^1$. $x=-1$ and $x=5$ both bounce — even, $(x+1)^2(x-5)^2$.

Coefficient, from the horizontal asymptote. $y=-2$. The negative already flipped into $(3-x)^3$, so the loose number out front is just $2$.

Degree check. Top: $1+2+2=5$. Bottom: $2+3=5$. Equal — confirmed.

$$f(x)=\dfrac{2(x+4)(x+1)^2(x-5)^2}{(x+3)^2(3-x)^3}$$

This was a multiple-choice problem with seven answer choices that all shared this exact numerator — only the denominator and the leading number changed between them. Building the equation yourself is faster than testing all seven.

The method

The Flow Chart — Same Order, Every Time

1 · Verticals → denominator. Every vertical asymptote is a factor in the bottom. Watch for a hidden one — if the graph never touches the $y$-axis anywhere, there's an asymptote at $x=0$ that nobody labeled for you.

2 · Zeros → numerator, cross or bounce. Straight through → odd power. Bounces back → even power. The flatter the bounce, the higher the power — but on this exam it's almost always a 2, 3, or 4.

3 · Horizontal asymptote → coefficient. That value is the ratio of leading coefficients. If it's negative, the sign has to live somewhere — either as a loose negative out front, or absorbed by flipping one odd-power factor's sign, e.g. $(x-2)\to(2-x)$.

4 · Degree check → confirm or eliminate. A horizontal asymptote that's a real, nonzero number only happens when the degree on top equals the degree on bottom. Add up every exponent — a bare $x$ by itself still counts as degree 1 — and check they match.

"It's a lot, but it's a flow chart. It's just a this, then this, then this." Your words, from today — after the second problem, not the first. That's normal. Keep running it in the same order until it's boring.

Your shaky spot extra reps

Cross or Bounce?

This is the one piece that's still catching you, so it gets its own drill — separated from the full graph problems below. No denominators, no coefficients, just: what does the graph do at this zero, and what does that tell you about the exponent?

1 · At $x=2$, the graph flattens against the axis and turns back the same direction it came from.

Cross or bounce? ›

Bounce — even. Simplest case, exponent $2$: factor $(x-2)^2$.

2 · At $x=-1$, the graph passes straight through, no hesitation.

Cross or bounce? ›

Cross — odd. Simplest case, exponent $1$: factor $(x+1)^1$.

3 · At $x=0$, the graph goes almost flat against the axis for a stretch, more than a normal bounce — but still turns back and doesn't cross.

Cross or bounce? ›

Still a bounce — even, just a higher one. The flatter it lies against the axis, the higher the power: could be $4$ instead of $2$. Still even, still doesn't cross. Exam answer choices are almost always $2$ or $4$ here.

4 · At $x=5$, the graph pauses and flattens briefly right at the axis, then keeps going through to the other side.

Cross or bounce? ›

Still crosses — odd, just a higher one. A brief flatten-then-continue is an odd power higher than 1, usually $3$. The test that never changes: does it end up on the other side of the axis or not? Ends up on the other side → odd, no matter how it hesitates on the way.

Worked example

Both Traps at Once

A hidden vertical asymptote and a sign that needs somewhere to live — in the same problem. Read the graph before you read the steps.

1 · Verticals. The curve never touches the $y$-axis anywhere on this graph — that's your tell. Hidden asymptote at $x=0$. There's also a visible one at $x=5$.

2 · Which is odd, which is even? At $x=0$, both branches shoot the same direction — even, $x^2$. At $x=5$, the branches go opposite directions — odd, and since $x=5$ is where the sign work happens, watch step 4.

3 · Zeros. $x=-2$ crosses — odd, $(x+2)^1$. $x=3$ bounces — even, $(x-3)^2$.

4 · Horizontal asymptote. $y=-2$, negative. The $x=5$ factor is the odd one, so the sign gets absorbed there: $(5-x)$, not $(x-5)$. That leaves a clean, positive $2$ out front.

5 · Degree check. Top: $1+2=3$. Bottom: $2+1=3$. Equal — confirmed.

$$f(x)=\dfrac{2(x+2)(x-3)^2}{x^2(5-x)}$$

Your turn — partly filled in

Faded Example — Finish Steps 4 and 5

Steps 1–3 are done for you below. This time the horizontal asymptote is positive — figure out whether the flip trick is even needed before you use it.

1 · Verticals (done). Hidden at $x=0$, visible at $x=-4$: denominator so far, $x^{\,?}(x+4)^{\,?}$.

2 · Zeros (done). $x=-3$ bounces — even. $x=1$ crosses — odd: numerator so far, $(x-1)^1(x+3)^2$.

3 · Which VA is even, which is odd? Look at the branches around each dashed line on the graph above and decide before you reveal.

Finish it — steps 3, 4, 5 ›

Step 3. At $x=0$, both sides shoot the same direction — even, $x^2$. At $x=-4$, the sides go opposite directions — odd, $(x+4)^1$.

Step 4 — the trap. $y=3$ is positive. There's nothing to flip. Even though $(x+4)$ is an odd-power factor, you only flip a factor when the sign forces it. Here it just stays $(x+4)$, and the coefficient sits out front as a plain $3$.

Step 5. Top: $1+2=3$. Bottom: $2+1=3$. Equal — confirmed.

$$f(x)=\dfrac{3(x-1)(x+3)^2}{x^2(x+4)}$$

Now you

Practice — One Feature at a Time

Each pair below is almost the same graph — only one thing changes. Spot what changed and why the equation changed with it. Stuck? Open a hint before the full reveal.

Pair 1 — does a hidden asymptote change anything else?

A — hidden VA at $x=0$

B — no hidden VA, both visible

Hint 1 ›

Same zero, same second asymptote. What's genuinely different about A?

Hint 2 ›

In A, the curve never touches the $y$-axis — that's the only clue you get for the asymptote at $x=0$. Nothing else about the method changes.

Full reveal ›

A: $\dfrac{x+1}{x(x-4)}$ B: $\dfrac{x+1}{(x-2)(x-4)}$ — same numerator, same method. The only change is one denominator factor, and A's is the one you'd miss if you only looked for x-intercepts near an asymptote. Both have degree $1$ on top, $2$ on bottom, so both have horizontal asymptote $y=0$ — a reminder that the degree-match rule only kicks in once the horizontal asymptote is a real, nonzero number.

Pair 2 — where does the negative sign go?

C — $\dfrac{x-2}{x-3}$

D — same graph, flipped

Hint 1 ›

Same zero, same asymptote location. What single number changed between C and D?

Hint 2 ›

Only the horizontal asymptote's sign flipped, from $+1$ to $-1$. Where can that new negative sign go?

Full reveal ›

D is $\dfrac{2-x}{x-3}$ — algebraically the same as $\dfrac{-(x-2)}{x-3}$. The negative from the horizontal asymptote got absorbed by flipping the numerator's odd-power factor instead of sitting out front. Same graph shape, same zero, same asymptote — only the sign moved house.

Capstone — every feature, and the sign hides where you don't expect

This time the hidden asymptote is the one that needs the flip, not the visible one. Don't assume the trick only applies to the obvious factor.

Hint 1 ›

Verticals first. Which one is hidden, and which one behaves the same on both sides (even) vs. flips (odd)?

Hint 2 ›

The hidden one at $x=0$ flips sign across it — that's the odd one this time. The visible one at $x=5$ goes the same direction both sides — even.

Hint 3 ›

Zeros: $-3$ bounces (even), $2$ crosses (odd). Horizontal asymptote is $-1$ — negative, and the odd-power factor is the hidden one, so that's where the flip happens.

Full reveal ›

Denominator: $x$ is odd/hidden and carries the flip → write it as part of $(2-x)$'s partner... more directly: $x(x-5)^2$ stays as-is, and the negative instead flips the numerator's odd factor: $(x-2)\to(2-x)$.

$$F(x)=\dfrac{(2-x)(x+3)^2}{x(x-5)^2}$$

Degree check: top $1+2=3$, bottom $1+2=3$. Equal — confirmed. The lesson: check which factor is odd before you decide where the sign goes — it isn't always the visible asymptote.

Spot the error

This Student Made One Mistake — Find It

1 · Graph: zeros at $x=-3$ (crosses) and $x=1$ (bounces), vertical asymptotes at $x=-1$ and $x=4$, horizontal asymptote $y=2$.

Student's work: "Numerator degree: $(x+3)$ is degree 1, $(x-1)^2$ is degree 2, total 3. Denominator: $(x+1)(x-4)$, degree 2. $3\ne 2$... but the horizontal asymptote is $2$, so it must still be fine."

What went wrong? ›

The degree mismatch is real, and it's not "still fine" — a nonzero horizontal asymptote requires equal degrees, full stop. This student found a genuine contradiction and then talked themselves out of trusting it. If the degrees don't match, this can't be the right reconstruction; go back and recheck a multiplicity before assuming the rule bends.

2 · Function pieces: numerator $x(x+5)$, denominator $(x-1)(x+3)$.

Student's work: "Numerator degree: $(x+5)$ is degree 1. Denominator degree: 2. Not equal, so this graph can't have a horizontal asymptote."

What went wrong? ›

They forgot to count the standalone $x$ factor. Numerator is really $x^1(x+5)^1$ — degree $2$, not $1$. With denominator degree $2$, the degrees actually do match. A bare $x$ sitting by itself is still a factor of degree 1 — count it every time.

Review no scaffolding — cold recall

Mixed Review

No graph is annotated for you here — some aren't drawn at all. Work from memory. Mixes today's skill with the asymptote material from the day before.

1 · A graph has zeros at $x=-2$ (crossing) and $x=4$ (bouncing), a vertical asymptote at $x=1$, and never touches the $y$-axis anywhere. The far ends of the graph flatten toward $y=3$. Write the full equation.

Check ›

$f(x)=\dfrac{3(x+2)(x-4)^2}{x(x-1)}$ — wait, check the degrees: top is $1+2=3$, bottom (with the hidden $x$) is $1+1=2$. They don't match a nonzero HA of $3$ — so a $2$nd power is needed on the denominator to make it work, e.g. $\dfrac{3(x+2)(x-4)^2}{x(x-1)^2}$ (degree $3$ over $3$). The point of this one: build it, then run the degree check yourself before trusting the answer.

2 · Explain in your own words why a horizontal asymptote that's a nonzero number forces the degrees on top and bottom to match — and what mistake catches people on the degree count.

Check ›

The horizontal asymptote comes from the ratio of leading terms as $x\to\pm\infty$. If the degrees didn't match, that ratio would either blow up (top bigger) or collapse to $0$ (bottom bigger) — the only way it settles on a real, nonzero number is if the degrees are equal, and the value is the ratio of leading coefficients. The classic slip: forgetting a standalone $x$ counts as degree 1.

3 · A rational function has a vertical asymptote at $x=1$ in one version, but a hole at $x=1$ in a slightly different version. What's the algebraic difference between those two functions?

Check ›

A hole happens when the same factor, like $(x-1)$, appears in both the numerator and denominator and cancels. An asymptote happens when the factor is only in the denominator. Same-looking graph near that point, completely different reason.

4 · Finish the mnemonic from earlier this week: "Bobo Voza eats DC." What does each part mean, in your own words — not the rhyme?

Check ›

Bigger bottom (degree) → horizontal asymptote at $0$. Bigger top → no horizontal asymptote, there's a slant one instead. Equal degrees ("DC") → divide the coefficients to get the horizontal asymptote.

5 · Without doing the full division, describe how you'd find the slant asymptote of $\dfrac{x^2+3x-1}{x-2}$ — name the technique and your stopping point.

Check ›

Polynomial long division, $(x^2+3x-1)\div(x-2)$. You only need the quotient — stop as soon as you have it and drop the remainder. That quotient (a line) is the slant asymptote.

6 · You've built $f(x)=\dfrac{(x-1)(x+3)^2}{(x-4)(x+2)}$. Without fully solving it, describe how you'd set up $f(x)>0$ — what changes about how you use the zeros and the vertical asymptote compared to just graphing this function?

Check ›

Every zero and every vertical-asymptote location becomes a boundary point on a number line, same as before — but now you test each region for sign only, and any point from the denominator is always excluded with a parenthesis, no matter what the inequality sign is.

Before You Leave — Cheat-Sheet Line

One line, in your own words: "Verticals → bottom. Zeros → top, cross=odd/bounce=even. Horizontal asymptote → coefficient, negative flips a factor. Degrees must match if the HA isn't zero — count the bare x."

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