Which Of The Following Functions Is Graphed Below: Complete Guide

Which Function Is It? How to Identify a Graph When You’re Stuck

Ever stared at a squiggle on a math test and thought, “Which of the following functions is graphed below?Practically speaking, ” You’re not alone. Plus, the moment a curve pops up on a page and the multiple‑choice list looks like a foreign language, most of us freeze. The short version is: you can train yourself to read a graph the way you read a paragraph—by looking for clues, patterns, and the occasional “aha!” moment.

Below is a step‑by‑step guide that takes the mystery out of those “which function?” questions. I’ll walk you through what the graph is actually showing, why it matters for your class (or just for satisfying your curiosity), the common pitfalls that trip up even seasoned students, and a handful of practical tricks you can use right now. By the end, you’ll be the kind of person who can glance at a curve and whisper, “That’s a cubic with a vertical shift,” without breaking a sweat.

What Is “Which Function Is Graphed Below?”

In plain English, the question asks you to match a visual representation—a curve on a coordinate plane—to its algebraic formula from a list. The list might include linear, quadratic, exponential, logarithmic, trigonometric, or even piecewise functions Easy to understand, harder to ignore. Practical, not theoretical..

The visual language of graphs

A graph isn’t just a pretty picture; it’s a compact story about how y behaves as x changes. Things to watch for:

• Intercepts – where the line crosses the axes.
• Symmetry – does it mirror across the y‑axis (even) or origin (odd)?
• Asymptotes – invisible walls the curve never crosses.
• Growth/decay – does it shoot upward, flatten out, or bounce?
• Periodicity – repeating patterns hint at sine or cosine.

When you see a graph, you’re really seeing a collection of these features stacked together. The challenge is to decode which combination matches one of the answer choices.

Why It Matters

First, the obvious: getting the right answer can be the difference between an A and a D on a quiz. But beyond grades, understanding how to read a graph builds a mental model that’s useful everywhere—from interpreting stock charts to spotting trends in a health study.

If you skip this skill, you’ll keep treating graphs as “just pictures” and miss out on the deeper insight they provide. Real‑world data rarely comes with a neat formula; you have to infer the shape first, then decide what model fits.

No fluff here — just what actually works Most people skip this — try not to..

How to Do It: A Systematic Approach

Below is the workflow I use every time I’m faced with a “which function?” problem. Feel free to tweak it; the goal is to make the process second nature.

1. Scan for Intercepts

• X‑intercept(s): Where does the curve hit the x‑axis? Count them.
• Y‑intercept: Plug x = 0 into the graph mentally; does it start at (0, 0), (0, 2), or somewhere else?

Why it helps:

• A linear function has at most one x‑intercept.
• A quadratic opens upward or downward and can have 0, 1, or 2 x‑intercepts.
• Exponential functions never cross the x‑axis (they have a horizontal asymptote at y = 0).

2. Look for Symmetry

• Even symmetry (mirror across the y‑axis) → likely a quadratic, cosine, or even‑powered polynomial.
• Odd symmetry (rotate 180° around the origin) → suggests a cubic, sine, or odd‑powered polynomial.

If you can’t spot perfect symmetry, note whether the left side looks like a stretched or flipped version of the right side And that's really what it comes down to..

3. Spot Asymptotes

• Vertical asymptote at x = a → rational function (e.g., 1/(x‑a)).
• Horizontal asymptote at y = b → exponential decay/growth or rational function where numerator degree < denominator degree.

4. Check End Behavior

Follow the curve far out to the left and right:

• Both ends up → even‑degree polynomial with positive leading coefficient (e.g., x⁴).
• Both ends down → even‑degree polynomial with negative leading coefficient.
• Left up, right down → odd‑degree polynomial with negative leading coefficient (e.g., ‑x³).
• Left down, right up → odd‑degree polynomial with positive leading coefficient (e.g., x³).

5. Identify Any Repeating Patterns

If the graph wiggles regularly, you’re looking at a trigonometric function. Note the distance between peaks (the period) and the height of the peaks (the amplitude).

6. Compare to Answer Choices

Now line up the clues you gathered with the list of functions. Eliminate anything that fails a single test—if a choice has a vertical asymptote but the graph shows none, cross it out.

7. Verify with a Quick Plug‑In

If you’re still torn between two possibilities, pick a convenient x‑value (like 1 or 2) and see which function gives a y‑value that matches the graph’s point Small thing, real impact..

Common Mistakes / What Most People Get Wrong

1. Relying on a single feature – “It looks like a parabola, so it must be quadratic.” Forgetting about asymptotes can lead you astray.

2. Ignoring scale – The grid spacing can exaggerate steepness. A shallow exponential can look linear if the axis is stretched.

3. Assuming symmetry when it’s only approximate – Real data often has noise; a curve may look symmetric but actually isn’t Simple as that..

4. Skipping the y‑intercept – Many students jump straight to x‑intercepts and miss the clue that the graph starts at (0, ‑3), for example, which points to a vertical shift.

5. Over‑trusting the calculator – Plotting a function on a cheap graphing app can introduce rounding errors that make a rational function look smooth Worth knowing..

Practical Tips / What Actually Works

• Keep a cheat sheet of “signature traits.” Write down one‑line notes like “exponential: never touches x‑axis, rapid growth after a flat start.”

• Use the “two‑point test.” Pick any two visible points, compute the slope or ratio, and see which function family fits.

• Draw a quick sketch of each answer choice on a spare piece of paper. Seeing the shapes side‑by‑side cements the differences.

• Remember the “shift” trick. If the whole graph is moved up or down, just subtract that constant from the y‑values you read off. It often turns a messy-looking curve into a textbook example.

• Practice with real data sets. Grab a CSV of temperature over a year, plot it, and try to name the function. The more contexts you see, the sharper your intuition becomes.

FAQ

Q: What if the graph looks like a piecewise function but the answer list only has single formulas?
A: Check for sharp corners or jumps. If you see them, the correct answer is probably a piecewise definition, or the test expects you to pick the “closest” single formula and note the discrepancy.

Q: How do I differentiate between a cubic and a quartic when both have three turning points?
A: Look at the end behavior. Cubics go opposite directions at the ends; quartics go the same way.

Q: The graph has a horizontal asymptote at y = 2, but one of the choices is y = 2 + eˣ. Is that a match?
A: No. eˣ grows without bound, so the curve would eventually leave the asymptote. A horizontal asymptote at y = 2 suggests a rational function where the numerator degree ≤ denominator degree.

Q: Can I use the derivative to identify the function?
A: In theory, yes—if you can estimate slopes from the graph. In practice, for a quick test it’s faster to rely on intercepts, symmetry, and asymptotes.

Q: What if the graph is blurry or low‑resolution?
A: Zoom in, redraw the key points on graph paper, and focus on the most distinctive features (like a single x‑intercept or a clear asymptote).

That’s it. Next time you see a curve and a list of formulas, you’ll have a solid game plan. No more guessing, just a systematic read‑the‑graph routine that turns a confusing picture into a clear answer. Happy graph‑hunting!

6. When the Graph Lies Somewhere Between Two Families

Sometimes a curve will display characteristics of two different families—say, it looks almost exponential but flattens out like a rational function. In those “borderline” cases, follow this decision tree:

1. Identify the end behavior first.
   If the y‑values keep climbing (or diving) without bound, you’re in the exponential/polynomial camp.
   If the curve levels off to a finite value on one or both sides, you’re looking at a rational or logarithmic function.

2. Check for a “knee.”
   A sharp bend that transitions from a gentle slope to a steep climb is a hallmark of an exponential with a horizontal asymptote (e.g., (y = a + be^{cx})). A gradual bend that never becomes vertical usually points to a rational function of the form (\frac{p(x)}{q(x)}) where (\deg p = \deg q) Small thing, real impact..

3. Count the asymptotes.
   One vertical asymptote + one horizontal asymptote → classic rational.
   Only a horizontal asymptote → exponential or logarithmic The details matter here..

4. Look for “inflection points.”
   Exponential curves have no inflection point (the curvature never changes sign). Rational functions often have one (where the graph switches from concave up to concave down). If you can spot a change in concavity, you’ve likely found a rational function.

5. Test a simple transformation.
   Subtract the suspected horizontal asymptote from the y‑values you read off the graph. If the resulting points line up roughly on a straight line when plotted on a semi‑log scale, the original curve is exponential. If they line up on a log‑log scale, you’re dealing with a power‑law rational function Most people skip this — try not to..

By walking through these steps quickly—most of which can be done mentally or with a scrap of paper—you can resolve the ambiguity without second‑guessing.

7. A Mini‑Checklist for the Test‑Taker

Short version: it depends. Long version — keep reading.

Keep this table in the margin of your notebook; a quick glance can eliminate half the answer choices instantly Easy to understand, harder to ignore..

8. Putting It All Together – A Walk‑Through Example

Imagine you’re given the following graph (a stylized “S” that flattens out near y = 3 on the right and dives toward –∞ as x → 0⁺). The answer list contains:

1. (y = \frac{5}{x+2} + 3)
2. (y = 3e^{2x})
3. (y = 3 - 5e^{-x})
4. (y = \frac{3x}{x+1})

Step 1 – Asymptotes.
The curve approaches y = 3 on the right → horizontal asymptote at 3. It also shoots up near x = 0 → vertical asymptote at x = 0.

Step 2 – Match candidates.
Only (1) and (3) have a horizontal asymptote at 3. (1) has a vertical asymptote at x = –2, not at 0, so discard it. (3) has no vertical asymptote at all, but the exponential term (e^{-x}) decays rapidly, never blowing up. The only remaining choice that satisfies both asymptotes is none of the above—so we double‑check the graph.

Step 3 – Re‑examine the graph.
A closer look shows the “blow‑up” occurs at x = –2, not at 0. The earlier impression was a visual artifact. Now (1) fits perfectly: horizontal asymptote y = 3, vertical asymptote x = –2, and the curve is decreasing, matching the negative sign in the numerator.

Conclusion: The correct answer is (1) (y = \frac{5}{x+2} + 3).

This short exercise demonstrates how the checklist and asymptote focus can turn a seemingly ambiguous picture into a decisive answer in under a minute Small thing, real impact..

Final Thoughts

Reading a graph is less about “guesswork” and more about a disciplined visual audit. Plus, by training yourself to spot intercepts, symmetry, asymptotes, and curvature first, you create a mental scaffold that instantly narrows the field of possible formulas. The cheat‑sheet, two‑point test, and quick sketching tricks are not optional extras—they’re the shortcuts that seasoned test‑takers use to shave precious seconds off each question.

Remember:

• Don’t chase the exact shape; chase the defining features.
• Use transformations (shifts, reflections, scalings) as a mental “undo” button to reduce a complex curve to a familiar template.
• Practice with real‑world data to cement the intuition that a rising temperature curve, a decaying radioactive‑sample curve, and a rational‑function growth curve all speak the same visual language.

With these habits in place, the next time a curve and a list of formulas appear together, you’ll approach it with confidence, not confusion. The graph will no longer be a mystery—it will be a map, and you’ll have the coordinates to figure out it straight to the correct answer Nothing fancy..

Happy graph‑reading, and may every curve you meet lead you directly to the right function!

Putting It All Together – A Walk‑Through of a Full‑Length Practice Question

Below is a representative GRE‑style “match the graph to the equation” problem. We’ll solve it step‑by‑step, applying every shortcut discussed so far Nothing fancy..

The Question

You are shown a graph of a curve that:

• passes through the point ((0,,2)) and ((2,,5))
• has a vertical asymptote at (x = -1)
• approaches a horizontal asymptote (y = 3) as (x \to \infty)
• is decreasing for (x < -1) and increasing for (x > -1)

Which of the following equations could generate this graph?

Step 1 – Write Down the “signature” of the graph

Step 2 – Eliminate choices quickly

Only A and B have the correct vertical asymptote at (x=-1). The horizontal asymptote for both is (y=3), matching the graph. The sign of the numerator decides the left‑hand behavior:

• With +5 (choice A) the fraction is positive on both sides of the asymptote, so the curve would be above the horizontal line (y=3) on both sides—contradicting the observed decreasing‑to‑(-\infty) left side.
• With –5 (choice B) the fraction is negative for (x>-1) (making the graph approach 3 from above) and also negative for (x<-1) (making it plunge to (-\infty) as we approach the asymptote from the left). This matches the described behavior perfectly.

Thus the only viable candidate is choice B And that's really what it comes down to..

Step 3 – Verify with the given point

Using choice B:

[ y = \frac{-5}{x+1}+3 ]

Plug (x=0):

[ y = \frac{-5}{1}+3 = -5+3 = -2 ]

But the graph passes through ((0,2)). Something is still off.

What went wrong? In many GRE items the curve is translated vertically and horizontally. The list of answer choices we were given omitted a possible constant term other than 3. The correct generic form is

[ y = \frac{k}{x+1}+c ]

where (c) is the horizontal asymptote. The problem statement said the asymptote is (y=3), so (c=3). The point ((0,2)) forces

[ 2 = \frac{k}{1}+3 \quad\Longrightarrow\quad k = -1. ]

Hence the actual equation that satisfies every condition is

[ \boxed{y = \frac{-1}{x+1}+3 }. ]

None of the four listed options contains (-1) in the numerator, so the correct answer—again—is “none of the above.”

What This Example Shows

1. Start with the must‑haves (asymptotes, intercepts). They whittle the list down dramatically.
2. Use the sign of the numerator to decide which side of the asymptote the curve lives on.
3. Plug a single point to nail down the exact constant in the numerator; if the resulting value isn’t among the answer choices, you’ve identified a “none of the above” situation.
4. Don’t be fooled by a second point that may have been mis‑copied; the asymptotes are far more reliable than an isolated interior point.

Quick‑Reference Cheat Sheet (One Page)

Keep this sheet in the margin of your scratch paper; a quick glance will often let you eliminate three‑quarters of the answer choices in less than ten seconds.

The Bottom Line

Graph‑matching questions on the GRE are a test of visual logic, not of artistic interpretation. By:

• Identifying asymptotes first,
• Marking intercepts and symmetry,
• Applying a two‑point (or one‑point + asymptote) test, and
• Using the transformation “undo” trick to reduce a complex curve to a basic template,

you turn a seemingly daunting picture into a series of algebraic clues that point to a single answer—sometimes “none of the above.”

Practice these steps on a handful of timed drills, and you’ll find that the time you spend on each graph drops from a minute to a few seconds, freeing up precious minutes for the more calculation‑heavy sections of the test Less friction, more output..

Happy graph‑reading, and may every curve you encounter guide you straight to the right function!

5. When the Graph Shows a “Break” or a Hole

A hole (removable discontinuity) is easy to miss because the curve looks continuous except for a tiny gap. On the GRE it almost always appears as a small open circle on an otherwise smooth line Nothing fancy..

How to encode a hole:

1. Write the unsimplified rational form: (\displaystyle f(x)=\frac{(x‑a)·N(x)}{(x‑a)·D(x)}).
2. Cancel ((x‑a)) to get the simplified function (g(x)=\frac{N(x)}{D(x)}).
3. The hole’s coordinate is ((a,,g(a))).

When you see a hole at, say, ((2,‑3)), you can immediately write a candidate numerator factor ((x‑2)) and a denominator factor ((x‑2)). The remaining part of the function must evaluate to (-3) at (x=2); that gives you a quick equation for any unknown constant.

6. A Shortcut for Quadratic‑Denominator Rational Functions

Many GRE items involve a “U‑shaped” curve that never touches the x‑axis but has a clear horizontal asymptote. Those are typically of the form

[ f(x)=\frac{a}{(x‑h)^2+k}+c, ]

where

• (c) is the horizontal asymptote,
• ((h,,c)) is the vertex of the “bowl,”
• (k>0) determines how “tight” the bowl is, and
• (a) controls whether the curve sits above ((a>0)) or below ((a<0)) the asymptote.

Reading the graph:

Typical test‑taking move: Identify (c) from the asymptote, read the vertex ((h, y_v)), and then solve for the ratio (\frac{a}{k}=y_v-c). Because the answer choices never require you to isolate both (a) and (k) separately, the ratio is enough to pick the correct option.

7. Putting It All Together – A Mini‑Walkthrough

Suppose the GRE presents a graph that looks like this:

• Horizontal asymptote at (y=2).
• A single hole at ((1,,3)).
• The curve passes through ((-2,,4)).
• The curve is decreasing for (x<1) and increasing for (x>1).

Step 1 – Write the skeleton.
The hole tells us there is a factor ((x‑1)) in both numerator and denominator. The horizontal asymptote (y=2) suggests the simplified form approaches 2, so we start with

[ f(x)=2+\frac{A}{(x‑1)}\quad\text{(unsimplified)}. ]

But we need a hole, not a vertical asymptote, so we multiply numerator and denominator by ((x‑1)):

[ f(x)=\frac{2(x‑1)+(A)}{x‑1}=2+\frac{A}{x‑1}. ]

Now cancel the ((x‑1)) factor to get the simplified function

[ g(x)=2+\frac{A}{x‑1}. ]

Step 2 – Use the hole’s coordinates.
At the hole, the original function is undefined, yet the simplified function would give the y‑value. Hence

[ g(1)=2+\frac{A}{0}\quad\text{(undefined)}. ]

Because the graph shows a hole at ((1,3)), the limit as (x\to1) must equal 3. Therefore

[ \lim_{x\to1}g(x)=3;\Longrightarrow;2+\frac{A}{0}=3, ]

which tells us the sign of (A): the fraction must approach (+1) from either side, so (A) must be positive and of magnitude 1. In practice we set

[ \frac{A}{x‑1}\to1;\text{as }x\to1;\Longrightarrow;A=1. ]

Step 3 – Verify with the known point ((-2,4)).

[ f(-2)=2+\frac{1}{-2‑1}=2-\frac13=\frac{5}{3}\neq4. ]

Our current skeleton is wrong; we missed a scaling factor. Let’s introduce a constant (k) in the denominator:

[ f(x)=2+\frac{A}{k(x‑1)}. ]

Now the limit as (x\to1) is still (2+\frac{A}{0}=3) → (A/k=1). So (A=k). Use the point ((-2,4)):

[ 4=2+\frac{k}{k(-2‑1)}=2+\frac{1}{-3}=2-\frac13=\frac{5}{3}. ]

Again a mismatch—clearly we need a quadratic denominator to get the correct shape. Switch to the quadratic‑denominator template:

[ f(x)=2+\frac{A}{(x‑1)^2+B}. ]

The hole forces a factor ((x‑1)) in both numerator and denominator, so the unsimplified version is

[ f(x)=\frac{2[(x‑1)^2+B]+A(x‑1)}{(x‑1)^2+B}. ]

Cancel the ((x‑1)) factor (the numerator contains (A(x‑1))), leaving the simplified form

[ g(x)=2+\frac{A}{(x‑1)^2+B}. ]

Now apply the hole: as (x\to1), the denominator → (B), so

[ \lim_{x\to1}g(x)=2+\frac{A}{B}=3;\Longrightarrow;\frac{A}{B}=1. ]

Thus (A=B). Use the known point:

[ 4=2+\frac{A}{(‑2‑1)^2+A}=2+\frac{A}{9+A}. ]

Solve:

[ 2=\frac{A}{9+A};\Longrightarrow;2(9+A)=A;\Longrightarrow;18+2A=A;\Longrightarrow;A=-18. ]

Since (A=B), we have (B=-18). The denominator ((x‑1)^2-18) indeed changes sign, giving the observed decreasing‑then‑increasing shape while preserving the hole at (x=1). The final function (unsimplified) is

[ \boxed{f(x)=\frac{2[(x‑1)^2-18]-18(x‑1)}{(x‑1)^2-18}}. ]

All answer choices that do not reduce to this expression can be eliminated instantly Not complicated — just consistent..

Conclusion

Graph‑matching questions on the GRE are less about memorizing a catalog of curve families and more about recognizing a handful of signature features—asymptotes, intercepts, holes, and symmetry—and then translating those visual cues into a compact algebraic skeleton. By:

1. Zeroing in on asymptotes first (they dictate denominator structure),
2. Marking every intercept and the location of any hole,
3. Using the “two‑point + asymptote” rule to solve for the remaining constants, and
4. Employing the quick‑reference cheat sheet to keep the process systematic,

you can reduce any GRE graph‑matching item to a short set of equations that point to a single answer—or to “none of the above.”

Practice the workflow on a few timed examples, keep the cheat sheet handy, and you’ll find that what once felt like a visual puzzle becomes a routine, high‑accuracy step in your test‑taking arsenal. Good luck, and may every curve you encounter lead you straight to the correct function!