From an Initial Condition to a Point
An initial condition such as \(y(2)=3\) gives a value of a function at a specified input. In a graph, that information is a location: when \(x=2\), the curve has height \(y=3\). The point is written \((2,3)\), with the input first and the output second.
When a slope field is accompanied by plotted solution curves, the curves show possible graphs of solutions. To read an initial condition from that picture, first mark its coordinate point, then find which plotted curve passes through that point. The direction marks around the curve indicate slopes; they do not change the coordinates in the initial condition.
This is a graph-reading task, not a request to solve the differential equation. The graph already displays the candidate curves. The task is to match the given point to the curve that goes through it.
Why the Point Is \((x_0,y_0)\)
A graph of a function \(y=f(x)\) consists of the points \((x,f(x))\): for each input, the graph places the function's output at that horizontal position. If \(f(x_0)=y_0\), then the graph includes the point with horizontal coordinate \(x_0\) and vertical coordinate \(y_0\).
This establishes the reading rule directly: an initial condition identifies a point on the solution curve because the graph records the value of the function at that input. The order matters. The first coordinate tells you where to look horizontally; the second tells you the height at that location.
A Reliable Graph-Reading Process
For a plotted slope field and its solution curves, follow the same coordinate steps each time. A vertical line through \(x=x_0\) helps you check the height of each candidate curve at the required input.
In \(y(x_0)=y_0\), identify the input \(x_0\) and output \(y_0\).
Write the point in the order \((x_0,y_0)\), and locate it using the graph's axis scales.
At \(x=x_0\), compare each curve's height with \(y_0\). The curve that passes through the point is the match.
Confirm that the selected curve is at the stated input and stated output, not merely close to one of them.
Reading both coordinates is important. A curve may pass through the correct height at a different input, or through the correct input at a different height. Neither is a match.
Worked Examples
Worked Example: Match a Curve to \(y(1)=2\)
A graph shows three candidate solution curves, labeled A, B, and C. On the vertical line \(x=1\), curve A passes through height \(y=1\), curve B passes through height \(y=2\), and curve C passes through height \(y=3\). Which curve matches the initial condition \(y(1)=2\)?
Step 1: Read the condition. The input is \(x_0=1\), and the output is \(y_0=2\).
Step 2: Write the point. In input-output order, the initial point is \((x_0,y_0)=(1,2)\).
Step 3: Compare the curves at the given input. At \(x=1\), A has height \(1\), B has height \(2\), and C has height \(3\). Only B has the required height.
Check: Curve B passes through \((1,2)\), so its value at input \(1\) is \(2\). Therefore, curve B matches \(y(1)=2\).
Worked Example: Read a Negative Input and a Fractional Output
A plotted set of curves is labeled R, S, and T. At \(x=-2\), their heights are \(y=2\) for R, \(y=1.5\) for S, and \(y=1\) for T. Which curve passes through the point specified by \(y(-2)=1.5\)?
Step 1: Read the condition. The input is \(x_0=-2\), and the output is \(y_0=1.5\).
Step 2: Form the coordinate pair. The point is \((x_0,y_0)=(-2,1.5)\). Move left from the vertical axis to \(x=-2\), then identify the height \(1.5\) on the vertical scale.
Step 3: Compare the ordinates. At \(x=-2\), R is at \(2\), S is at \(1.5\), and T is at \(1\). The required ordinate is \(1.5\), so S is the matching curve.
Check: S goes through \((-2,1.5)\), which means its value at input \(-2\) is \(1.5\). Thus S matches \(y(-2)=1.5\).
Worked Example: Check Both Coordinates at the Axes
Three curves, U, V, and W, are plotted near the axes. At \(x=0\), U has height \(-1.5\), V has height \(-1\), and W has height \(0\). Curve U also passes through \((1,-1)\). Which curve matches \(y(0)=-1\)?
Step 1: Identify the input and output. Here \(x_0=0\) and \(y_0=-1\).
Step 2: Write the point. The initial condition specifies \((0,-1)\): the point is on the vertical axis, one unit below the horizontal axis.
Step 3: Compare at \(x=0\). U is at \((0,-1.5)\), V is at \((0,-1)\), and W is at \((0,0)\). V is the only curve at the requested height when \(x=0\).
Check: V passes through \((0,-1)\), so \(y(0)=-1\) on V. U's point \((1,-1)\) has the correct height but the wrong input. Therefore, V is the answer.
The Initial Point Is Not the Slope
A slope field represents the derivative \(dy/dx\) at points in the plane. Its short marks indicate the direction a solution curve would have locally. An initial condition, by contrast, gives a location. These pieces of information are related in a slope-field graph, but they have different meanings.
| Information | What it tells you | How to read it |
|---|---|---|
| \(y(x_0)=y_0\) | The solution's value at the input \(x_0\) | The point \((x_0,y_0)\) |
| A slope mark at \((x,y)\) | The local derivative there | The direction or steepness of a short line segment |
| A plotted solution curve | How a candidate solution is drawn across the graph | Check whether it passes through \((x_0,y_0)\) |
It can help to keep two questions separate: “Where must the solution be?” is answered by the initial condition; “Which way is the solution directed nearby?” is indicated by the slope field. For this task, identify the required location first and use the drawn curves to find a match.
Common Mistake: Reversing the Coordinates
A frequent error is to treat \(y(1)=2\) as the point \((2,1)\). That reverses the input and output. The notation \(y(1)\) means “the value of \(y\) when \(x=1\),” so the horizontal coordinate is \(1\) and the vertical coordinate is \(2\).
Another mistake is to select a curve because it passes through the correct height somewhere on the graph. For example, a curve passing through \((1,-1)\) does not satisfy \(y(0)=-1\): it has height \(-1\) at input \(1\), not input \(0\). Always check both coordinates together.
When the Drawing Is Not Precise
Printed graphs and sketches have limited resolution. If two curves appear to pass through nearly the same location, inspect the grid, axis labels, and any stated coordinates rather than relying on the apparent thickness of the lines. A point halfway between labeled horizontal grid lines, for example, must be read using the graph's stated scale.
If no candidate curve visibly passes through the specified point, do not force a match. Recheck the coordinate order, signs, and scale. If those are correct and the drawing still does not show a clear match, the picture may be too imprecise to decide. The right conclusion is that the plotted evidence is insufficient for a confident identification.
This careful reading matters because a mistaken point leads to a mistaken curve, even when the graph itself is drawn clearly. On AP questions, writing the ordered pair before choosing a curve is a simple way to make the reasoning visible and reduce coordinate errors.
Check Your Understanding
For each question, identify the point specified by the initial condition before deciding which plotted curve would match.
- What point does \(y(3)=-2\) specify?
- A curve passes through \((-1,4)\). What initial condition does this point represent?
- At \(x=2\), curve A is at height \(0\), curve B at height \(2\), and curve C at height \(4\). Which curve matches \(y(2)=2\)?
- Why does a curve passing through \((1,5)\) not match \(y(5)=1\)?
- What should you do if the graph is too imprecise to tell whether a curve passes through the specified point?
