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Learn to solve quadratic equations by graphing parabolas and reading their x-intercepts. Includes worked example, common mistakes, rubric, and four practice tasks.
A quadratic equation in the form y = ax² + bx + c can be solved by graphing the parabola and finding where it crosses the x-axis. Those crossing points are the x-intercepts, and their x-values are the solutions (roots) of the equation. This lesson focuses only on graphing and reading intercepts; factoring and the quadratic formula are separate methods. You will build a table of values, plot points, draw a smooth curve, and record each x-intercept as a solution. When the parabola does not cross the x-axis, the equation has no real solutions. When it touches the axis at one point, there is one repeated real solution. Always label axes and use consistent units.
Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.
The graph of y = ax² + bx + c is a parabola. The solutions of ax² + bx + c = 0 are exactly the x-values where y = 0, which are the x-intercepts. To solve by graphing, plot enough points to see the curve clearly, then read the x-coordinates of the crossing points.
A parabola can cross the x-axis twice, touch it once, or miss it entirely. Two crossings mean two distinct real solutions; one touching point means one repeated real solution; no crossing means no real solutions. The y-intercept is at (0, c) and helps anchor the graph but is not a solution unless c = 0.
Choose x-values that bracket the likely turning point. For y = x² − 4, values from −3 to 3 give y = 5, 0, −3, −4, −3, 0, 5. The points (−2, 0) and (2, 0) lie on the x-axis, so the solutions are x = −2 and x = 2.
Plot at least five to seven points, including both sides of the vertex. Connect them with a smooth curve, not straight segments. Check symmetry: points equidistant from the vertex should have equal y-values. Label the x-axis and y-axis and mark each intercept clearly.
For y = x² + 1, every y-value is at least 1, so the parabola never reaches the x-axis; the equation x² + 1 = 0 has no real solutions. For y = x², the curve touches the x-axis only at (0, 0), giving one repeated solution x = 0.
Always state the number of solutions and their values. If the graph is unclear, add more points near the suspected intercept. A graphing convention: each grid square represents one unit unless labeled otherwise; state your scale before plotting.
Solve x² − 2x − 3 = 0 by graphing y = x² − 2x − 3.
The x-intercepts are x = −1 and x = 3, so the solutions are x = −1 and x = 3.
Solve x² − 4 = 0 by graphing y = x² − 4. Use x-values from −3 to 3.
x = −2 and x = 2.
Why: Table: x = −3 gives 5; −2 gives 0; −1 gives −3; 0 gives −4; 1 gives −3; 2 gives 0; 3 gives 5. The parabola crosses the x-axis at (−2, 0) and (2, 0).
Check: Substitute each solution: (−2)² − 4 = 0 and 2² − 4 = 0. Both equal zero.
Solve x² + 1 = 0 by graphing y = x² + 1. Use x-values from −2 to 2.
No real solutions.
Why: Table: x = −2 gives 5; −1 gives 2; 0 gives 1; 1 gives 2; 2 gives 5. All y-values are positive, so the parabola never meets the x-axis.
Check: Since x² ≥ 0 for all real x, x² + 1 ≥ 1 > 0, so y is never zero.
Solve x² = 0 by graphing y = x². Use x-values from −2 to 2.
x = 0 (one repeated real solution).
Why: Table: x = −2 gives 4; −1 gives 1; 0 gives 0; 1 gives 1; 2 gives 4. The parabola touches the x-axis only at (0, 0).
Check: Substitute x = 0: 0² = 0. The graph touches the axis at exactly one point.
Solve x² − 2x − 3 = 0 by graphing y = x² − 2x − 3. Use x-values from −2 to 4.
x = −1 and x = 3.
Why: Table: x = −2 gives 5; −1 gives 0; 0 gives −3; 1 gives −4; 2 gives −3; 3 gives 0; 4 gives 5. The parabola crosses the x-axis at (−1, 0) and (3, 0).
Check: Substitute: (−1)² − 2(−1) − 3 = 1 + 2 − 3 = 0; 3² − 2(3) − 3 = 9 − 6 − 3 = 0.
Then the equation has no real solutions; the graph never reaches y = 0.
At least five to seven points, including both sides of the vertex, so the curve is clear.
Yes, but add points near the intercept to confirm the value; state if it is an estimate.
No; graphing shows only real x-intercepts. Complex solutions are outside this method.