HomeGraphing Linear-Quadratic Systems: Find Intersection Points Free original lessons
Graphing Linear-Quadratic Systems: Find Intersection Points
Learn to graph a line and a parabola on the same axes, identify intersection points, and verify solutions algebraically with clear steps and practice.
Free lesson
Four exercises
Worked answers
Algebra and Function Practice
A linear-quadratic system pairs a line with a parabola. The solutions are the points where the two graphs cross. In this lesson you will plot both equations on one coordinate grid, read the intersection points, and confirm each point by substitution. The method works for any line and parabola, whether they meet twice, once, or not at all. You will also learn a consistent convention for sketching graphs so that your work can be checked fairly, even when handwriting and letter styles vary.
Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.
Set Up One Coordinate Grid
Choose a scale that fits both graphs. For example, if the parabola opens upward with vertex near the origin and the line has a moderate slope, a grid from negative five to five on both axes is often enough. Label the axes and mark equal spacing. Plot the line first because it is quick: find two points by substituting simple x-values, then draw a straight line through them. Next plot the parabola by making a small table of x-values around its vertex. Use the same scale for both graphs so intersections can be read accurately.
A useful convention for sketching: treat the baseline of your writing as the x-axis and the midline of lowercase letters as a reference height for labeling tick marks. Letter styles vary, so do not rely on a single handwriting shape; instead, keep tick labels aligned to the grid lines and write numbers clearly. If you use a ruler, measure your actual paper and pencil, not an assumed standard. Any numerical grid size you choose is a hypothetical example for that drawing, not a universal rule.
- Draw and label the x-axis and y-axis with equal tick spacing.
- Plot two points for the line and connect them with a straight edge.
- Make a table of at least five x-values for the parabola, including the vertex.
- Plot the parabola points smoothly and label each curve.
- Mark every crossing point and write its coordinates.
Read Intersection Points Carefully
An intersection point is where the line and parabola share the same x and y values. Read the coordinates from the grid by tracing down to the x-axis and across to the y-axis. If a crossing falls between grid lines, estimate to the nearest half unit and state that it is an estimate. A parabola can meet a line at two points, at one point (tangent), or at no real points. Always check how many crossings your graph shows before writing the solution set.
When the line is steep or the parabola is narrow, the crossing may be hard to read. In that case, use the graph to estimate, then verify by substitution. If substitution does not match, adjust your reading or replot with a finer scale. Keep units consistent: if x is in whole units, do not switch to tenths midway through the table.
- Circle each crossing point on the graph.
- Write the x-coordinate first, then the y-coordinate, as an ordered pair.
- Count the number of crossings to decide whether there are two, one, or no solutions.
- Substitute each ordered pair into both original equations to check.
- State the solution set as a list of ordered pairs.
Verify Algebraically
Graphing gives a visual answer, but substitution confirms it. Take an ordered pair (x, y) from your graph and substitute the x-value into both equations. The line equation should produce the y-value you read, and the parabola equation should produce the same y-value. For example, with the line y = x + 2 and the parabola y = x squared, the point (2, 4) checks because 2 + 2 = 4 and 2 squared = 4. The point (-1, 1) also checks because -1 + 2 = 1 and (-1) squared = 1. A point such as (1, 2) would not check: the line gives 1 + 2 = 3, not 2, and the parabola gives 1 squared = 1, not 2. Always compute both sides.
If a point fails, recheck your table and plotting. A common cause is using a different scale for the two curves. Another cause is misreading a negative sign. Keep your arithmetic simple and write each substitution on its own line so a partner can follow your reasoning.
- Write the ordered pair you read from the graph.
- Substitute the x-value into the line equation and compute y.
- Substitute the same x-value into the parabola equation and compute y.
- Compare both results with the y-value from the graph.
- If they match, include the point in the solution set; if not, recheck the graph.
Worked example
Graph the system y = x + 1 and y = x squared - 1 on the same axes. Find the intersection points and verify them.
Worked answer
The intersection points are (-1, 0) and (2, 3). Check (-1, 0): line gives -1 + 1 = 0; parabola gives (-1) squared - 1 = 0. Check (2, 3): line gives 2 + 1 = 3; parabola gives 2 squared - 1 = 3. Both points satisfy both equations.
Four practice exercises
Exercise 1
Graph y = x + 2 and y = x squared on the same axes. Find the intersection points and verify them.
Show answer and check
The intersection points are (-1, 1) and (2, 4).
Why: For the line, points (-1, 1) and (2, 4) lie on y = x + 2. For the parabola, (-1) squared = 1 and 2 squared = 4, so both points satisfy y = x squared. The graph shows two crossings.
Check: Substitute x = -1 into both equations: line gives 1, parabola gives 1. Substitute x = 2: line gives 4, parabola gives 4. Both match.
Exercise 2
Graph y = -x + 4 and y = x squared - 5x + 7 on the same axes. Find the intersection points and verify them.
Show answer and check
The intersection points are (1, 3) and (3, 1).
Why: Set the expressions equal: x squared - 5x + 7 = -x + 4. Add x and subtract 4 from both sides: x squared - 4x + 3 = 0. Factor: (x - 1)(x - 3) = 0, so x = 1 or x = 3. The line gives y = -1 + 4 = 3 at x = 1 and y = -3 + 4 = 1 at x = 3. The parabola gives 1 - 5 + 7 = 3 at x = 1 and 9 - 15 + 7 = 1 at x = 3. The graph shows two crossings, so there are two solutions.
Check: Substitute (1, 3): line -1 + 4 = 3; parabola 1 - 5 + 7 = 3. Substitute (3, 1): line -3 + 4 = 1; parabola 9 - 15 + 7 = 1. Both points satisfy both equations.
Exercise 3
Graph y = 2x - 3 and y = x squared - 4x + 5. Find the intersection points and verify them.
Show answer and check
The intersection points are (2, 1) and (4, 5).
Why: Set x squared - 4x + 5 = 2x - 3, so x squared - 6x + 8 = 0, which factors as (x - 2)(x - 4) = 0. Thus x = 2 or x = 4. The line gives y = 1 and y = 5. The parabola gives 4 - 8 + 5 = 1 and 16 - 16 + 5 = 5.
Check: Substitute (2, 1): line 2(2) - 3 = 1; parabola 4 - 8 + 5 = 1. Substitute (4, 5): line 8 - 3 = 5; parabola 16 - 16 + 5 = 5.
Exercise 4
Graph y = x - 1 and y = x squared - 2x + 1 on the same axes. How many intersection points are there? Find them and verify.
Show answer and check
There are two intersection points: (1, 0) and (2, 1).
Why: Set the expressions equal: x squared - 2x + 1 = x - 1. Subtract x and add 1 to both sides: x squared - 3x + 2 = 0. Factor: (x - 1)(x - 2) = 0, so x = 1 or x = 2. The line gives y = 0 at x = 1 and y = 1 at x = 2. The parabola gives 1 - 2 + 1 = 0 at x = 1 and 4 - 4 + 1 = 1 at x = 2. The graph shows two crossings, so there are two solutions.
Check: Substitute (1, 0): line 1 - 1 = 0; parabola 1 - 2 + 1 = 0. Substitute (2, 1): line 2 - 1 = 1; parabola 4 - 4 + 1 = 1. Both points satisfy both equations.
Common mistakes
- Using different scales on the x-axis and y-axis without noting it. Use equal tick spacing on both axes, or write the scale clearly beside each axis so intersections are read correctly.
- Plotting only three points for the parabola and drawing a straight line between them. Use at least five points around the vertex and draw a smooth curve, not a series of straight segments.
- Reading an intersection as (y, x) instead of (x, y). Trace down to the x-axis first, then across to the y-axis, and write the ordered pair in that order.
- Assuming every line and parabola cross twice. Count the crossings on your graph; a tangent line gives one point and a non-crossing line gives no real solutions.
Self-check rubric
- Axes are labeled and tick spacing is equal or clearly stated.
- The line is drawn through two correctly computed points.
- The parabola uses at least five points around the vertex and is drawn as a smooth curve.
- All intersection points are marked and written as ordered pairs (x, y).
- Each intersection point is verified by substitution into both equations.
- The number of solutions (two, one, or none) is stated consistently with the graph.
- Handwriting and letter styles may vary; credit is given for clear labels and consistent scale, not for a single style.
Questions and answers
How many solutions can a line and parabola have?
There can be two, one, or no real intersection points. The graph shows which case applies.
Do I have to verify every point algebraically?
Verification is strongly recommended because graph reading can be approximate. Substitute each ordered pair into both equations.
What if the intersection is between grid lines?
Estimate to the nearest half unit, state that it is an estimate, and use substitution to confirm the exact value if possible.
Can I use a different grid size?
Yes. Choose a scale that fits both graphs and label it clearly. Any numerical grid size is a choice for that drawing, not a fixed rule.