Graphing A Long Call At Expiration

What a Long Call Payoff Diagram Shows

A payoff diagram turns the long call’s expiration payoff into a picture. The horizontal axis records the underlying asset’s price at expiration, and the vertical axis records the call holder’s payoff at that same time. The exercise price, or strike price, determines where the graph changes from a horizontal zero-payoff segment to an upward-rising segment.

The diagram describes the option’s payoff, not the holder’s overall investment result. In particular, the graph below does not subtract the amount paid to acquire the option. Keeping the vertical axis clearly labelled as payoff helps prevent that distinction from being lost.

Definition: For one long call on one unit of an underlying asset, with exercise price $K$ and underlying price $S_T$ at expiration, the payoff per unit is $\max(S_T-K,0)$. Here $K\geq 0$ and $S_T\geq 0$. The diagram plots this payoff vertically against $S_T$ horizontally.

This definition assumes the call gives its holder the right, but not the obligation, to buy one unit of the underlying asset for $K$ at expiration. It concerns the payoff at expiration only. The formula does not describe the call’s value before expiration, when time remaining and other factors can matter.

Deriving the Shape from the Exercise Decision

At expiration, the holder compares the value of receiving the asset with the amount that must be paid to exercise. If $S_T>K$, exercising means paying $K$ for an asset worth $S_T$, producing a payoff of $S_T-K$. If $S_T\leq K$, exercising would not produce a positive payoff, so the holder can leave the call unexercised and receive zero.

These two cases give the piecewise formula:

\text{Long call payoff per unit} = \begin{cases} 0, & 0\leq S_T\leq K,\\ S_T-K, & S_T>K. \end{cases}

At $S_T=K$, the second expression would also equal zero. Including equality in the first case simply makes the boundary explicit. The maximum notation and the piecewise formula are equivalent: the payoff is whichever is greater, $S_T-K$ or zero.

\max(S_T-K,0)= \begin{cases} 0, & S_T\leq K,\\ S_T-K, & S_T>K. \end{cases}

The graph follows directly. For every underlying price from zero through $K$, plot a payoff of zero. Starting at the point $(K,0)$, plot positive payoffs that increase by the same amount as the underlying price increases. For example, if $S_T$ rises by one dollar while remaining above $K$, $S_T-K$ also rises by one dollar. Connect the points with a horizontal line up to the strike and a straight rising line beyond it.

Key takeaway: On a long call payoff diagram, label the horizontal axis $S_T$ and the vertical axis “payoff per unit.” The payoff is zero through $K$, including at $K$, and rises as $S_T-K$ for prices above $K$.

The following schematic shows the intended shape; it is not drawn to scale.

Long call payoff per unit
Payoff
↑
│　　　　　　　　　　　　　／
│　　　　　　　　　　　／
│　　　　　　　　　／
│＿＿＿＿＿＿＿＿●＿＿＿＿＿＿→ $S_T$
　　　　　　　　　 $K$

The horizontal segment lies on the zero-payoff axis. Its endpoint is the strike-price point $(K,0)$. The rising segment begins at that same point; it does not begin above the horizontal axis. The two segments meet at the boundary, so there is no jump in payoff there.

A Reliable Plotting Method

Before drawing, identify the quantity on each axis and the unit of the payoff. Then calculate values at the strike and at selected prices on both sides of it. At least one price below $K$, the boundary $K$ itself, and one price above $K$ make the shape easy to verify. Use a consistent scale on each axis and label the strike directly on the horizontal axis.

Write the inputs.
Record the exercise price $K$, the expiration price $S_T$ values to plot, and whether the payoff is per unit or for a specified number of units.

Calculate each payoff.
For every selected $S_T$, evaluate $\max(S_T-K,0)$. Do not subtract an option purchase price when the requested quantity is payoff.

Mark the strike boundary.
Plot $(K,0)$, then place zero-payoff points at prices below $K$ and positive-payoff points above $K$.

Connect and check.
Draw a horizontal segment through the zero-payoff region and a straight rising segment beyond $K$. Confirm that the graph never falls below zero.

Worked Examples

Worked Example: Plotting a Call with a $40 Strike

A hypothetical call gives its holder the right to buy one share for $K=\$40$ at expiration. Construct a set of points for a per-share payoff diagram using expiration prices of $\$0$, $\$30$, $\$40$, $\$50$, and $\$70$.

Method. Evaluate $\max(S_T-K,0)$ at each price, then plot the resulting ordered pairs $(S_T,\text{payoff})$.

At $S_T=\$0$, the payoff is $\max(0-40,0)=\max(-40,0)=\$0$. At $S_T=\$30$, it is $\max(30-40,0)=\max(-10,0)=\$0$. At the strike, it is $\max(40-40,0)=\$0$.

At $S_T=\$50$, the payoff is $\max(50-40,0)=\$10$. At $S_T=\$70$, it is $\max(70-40,0)=\$30$. The plotted points are therefore $(0,0)$, $(30,0)$, $(40,0)$, $(50,10)$, and $(70,30)$, with dollar amounts on both axes.

Draw the zero-payoff segment from $S_T=0$ through $S_T=40$. From $(40,0)$, draw the rising straight segment through $(50,10)$ and $(70,30)$. As a check, each plotted point above the strike has a payoff equal to its horizontal coordinate minus 40.

Common mistake avoided: Plotting a positive payoff at the strike would be incorrect. At $S_T=40$, buying for $\$40$ an asset worth $\$40$ produces no positive payoff.

Worked Example: Keeping Per-Share and Contract Payoff Units Consistent

In a hypothetical contract, each call represents 100 shares and has an exercise price of $K=\$25$ per share. An investor holds two such calls. Find the payoff per share and the combined payoff for the two contracts at expiration prices of $\$20$, $\$25$, $\$29$, and $\$34$. State which values belong on a per-share diagram.

Method. First calculate the payoff for one share using $\max(S_T-K,0)$. Then multiply by 100 shares per contract and by two contracts to find the combined contract payoff.

At $S_T=\$20$, the per-share payoff is $\max(20-25,0)=\$0$. The combined payoff is $0\times100\times2=\$0$. At $S_T=\$25$, the per-share payoff is $\max(25-25,0)=\$0$, so the combined payoff is also $\$0$.

At $S_T=\$29$, the per-share payoff is $\max(29-25,0)=\$4$. The combined payoff is $4\times100\times2=\$800$. At $S_T=\$34$, the per-share payoff is $\max(34-25,0)=\$9$, and the combined payoff is $9\times100\times2=\$1{,}800$.

For a per-share diagram, plot $(20,0)$, $(25,0)$, $(29,4)$, and $(34,9)$, with the strike marked at $\$25$. The combined values, $\$0$, $\$0$, $\$800$, and $\$1{,}800$, are correct for the two-contract position but use a different vertical scale and unit.

Check. At $\$34$, the price is $\$9$ above the strike. Two contracts represent $2\times100=200$ shares, so $200\times\$9=\$1{,}800$, as calculated. The common mistake is to plot the total contract payoff on a vertical axis labelled “per share.”

Worked Example: Reconstructing and Checking a Diagram

A hypothetical long call has a per-share payoff of $\$13$ when the underlying price at expiration is $\$58$. Its payoff at $S_T=\$45$ is zero. Determine the exercise price, give three points that anchor the graph, and check whether a proposed point $(\$63,\$18)$ is consistent.

Method. A positive call payoff means $S_T>K$, so the payoff equation is $S_T-K$. Use the given positive payoff to solve for $K$, then verify the zero-payoff region and proposed point.

Since the payoff at $\$58$ is $\$13$, $58-K=13$. Therefore $K=58-13=\$45$. The zero payoff given at $S_T=\$45$ agrees with the boundary value $\max(45-45,0)=\$0$.

Three useful points are $(\$40,\$0)$, $(\$45,\$0)$, and $(\$58,\$13)$. The first is below the strike, the second is the strike point, and the third is above the strike. At $S_T=\$63$, the formula gives $\max(63-45,0)=\$18$, so the proposed point $(\$63,\$18)$ is consistent.

Draw the horizontal segment through $\$45$ on the price axis, then the rising straight segment through the two positive-payoff points. The check also confirms that the proposed point is on the same rising segment, rather than on the zero-payoff portion.

Common mistake avoided: Do not treat the positive payoff of $\$13$ as the strike. It is the difference $S_T-K$; solving that difference gives the strike of $\$45$.

Common Diagram Errors

Most errors arise from reversing the role of the strike, mixing units, or graphing a quantity other than payoff. A quick formula check at one point below the strike, at the strike, and above the strike catches many of them.

Reversing the subtraction. The call payoff above the strike is $S_T-K$, not $K-S_T$. For $K=\$40$ and $S_T=\$50$, the correct payoff is $\$10$, not a negative amount.
Starting the rising segment at the wrong height. At $S_T=K$, the payoff is zero. The rising segment starts at $(K,0)$, not at $(K,K)$.
Drawing positive payoff below the strike. If $S_T<K$, then $S_T-K<0$, and the maximum with zero is zero. Do not extend the negative value onto the payoff axis.
Confusing payoff with a net result. The formula here measures the call’s expiration payoff. Subtracting an acquisition cost changes the quantity being graphed.
Changing units mid-graph. A payoff per share and a payoff for multiple contracts differ by the number of shares represented. Label the vertical axis to show which one is plotted.

These checks matter because the diagram communicates both the exercise-price threshold and how the expiration payoff varies on either side of it. A misplaced boundary or inconsistent scale can make a correct formula appear to describe a different contract.

Check Your Understanding

For each problem, identify the payoff values and the graph features requested. Payoffs are at expiration; unless stated otherwise, use dollars per underlying unit.

A call has $K=\$18$. Find the payoffs at $S_T=\$12$, $\$18$, and $\$23$, and state the boundary point on the diagram. (Answer: $\$0,\ \$0,\ \$5$; boundary $(\$18,\$0)$.)
A call has $K=\$36$. A proposed graph gives payoff $\$9$ at $S_T=\$45$ and zero at $S_T=\$36$. Verify both values and state which segment contains each point. (Answer: Both are correct; $(\$36,\$0)$ is the boundary and $(\$45,\$9)$ is on the rising segment.)
A call’s payoff is $\$7$ at $S_T=\$52$, and its payoff is zero at $S_T=\$44$. Find $K$, then calculate the payoff at $S_T=\$60$. (Answer: $K=\$45$; payoff at $\$60$ is $\$15$.)
A hypothetical position consists of three calls, each representing 100 shares, with $K=\$30$. At $S_T=\$38$, find the payoff per share and the combined payoff. If the graph’s vertical axis is labelled per share, which value belongs on it? (Answer: $\$8$ per share; $\$2{,}400$ combined; plot $\$8$ on the per-share graph.)
A candidate plots a long call with $K=\$24$ as a rising line from $(\$0,\$24)$ through $(\$24,\$0)$, then a horizontal zero segment above the strike. Identify the errors and give the correct payoff at $S_T=\$0$, $\$24$, and $\$31$. (Answer: The segments are reversed and the below-strike value is incorrectly positive; correct payoffs are $\$0,\ \$0,\ \$7$.)

What You'll Learn