Systems of a Line and a Parabola
Aligned to A-REI.C.7 — Common Core State Standards for Mathematics.
What this lesson teaches
Solve a system of a linear and a quadratic equation by substitution: solve the linear equation for one variable and substitute into the quadratic. The solutions are the points where the line meets the parabola.
Worked examples
Example 1
Solve y = x^2 and y = x + 2. Substitute to get x^2 = x + 2, so x^2 − x − 2 = 0, which factors as (x − 2)(x + 1) = 0, giving x = 2 or x = −1. The intersection points are (2, 4) and (−1, 1).
Example 2
Solve y = x² − 4x + 5 and y = 2x − 3. Substitute: x² − 4x + 5 = 2x − 3, so x² − 6x + 8 = 0, which factors as (x − 2)(x − 4) = 0. Then x = 2 gives y = 1, and x = 4 gives y = 5. The line cuts the parabola at (2, 1) and (4, 5).
Example 3
Common mistake: assuming a line and a parabola always meet twice. Solve y = x² + 1 with y = −2: substituting gives x² = −3, which has no real solution — the line passes entirely below the parabola. And a tangent line meets it exactly once, giving a repeated root. The discriminant of the combined equation tells you which case you are in: positive means two points, zero means one, negative means none.
Practice questions
- Solve y = x^2 and y = 4 for the intersection points.
- Solve y = x + 1 and y = x^2.
- Solve the system y = 2x − 1 and y = x^2 − 4 algebraically.
Watch the lesson
Every lesson comes with a video taught in English and Spanish — the same video the QR code in the printed workbook opens.
▶ Watch this lessonEn español
Sistemas de una recta y una parábola
Resuelve un sistema de una ecuación lineal y una cuadrática por sustitución: despeja una variable en la ecuación lineal y sustitúyela en la cuadrática. Las soluciones son los puntos donde la recta se encuentra con la parábola.
Ejemplo 1: Resuelve y = x^2 y y = x + 2. Sustituye para obtener x^2 = x + 2, así que x^2 − x − 2 = 0, que se factoriza como (x − 2)(x + 1) = 0, dando x = 2 o x = −1. Los puntos de intersección son (2, 4) y (−1, 1).
Ejemplo 2: Resuelve y = x² − 4x + 5 y y = 2x − 3. Sustituye: x² − 4x + 5 = 2x − 3, así que x² − 6x + 8 = 0, que se factoriza como (x − 2)(x − 4) = 0. Entonces x = 2 da y = 1, y x = 4 da y = 5. La recta corta la parábola en (2, 1) y (4, 5).
Ejemplo 3: Error común: suponer que una recta y una parábola siempre se cortan dos veces. Resuelve y = x² + 1 con y = −2: sustituir da x² = −3, que no tiene solución real — la recta pasa por completo debajo de la parábola. Y una recta tangente la toca exactamente una vez, dando una raíz repetida. El discriminante de la ecuación combinada dice en qué caso estás: positivo son dos puntos, cero es uno, negativo es ninguno.
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