CurriculumGrade 12Mathematics

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 example

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).

Practice questions

  1. Solve y = x^2 and y = 4 for the intersection points.
  2. Solve y = x + 1 and y = x^2.
  3. 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.

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En 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: 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).

More Algebra lessons

The Remainder TheoremA-APR.B.2Zeros of Polynomials and Their GraphsA-APR.B.3Proving Polynomial IdentitiesA-APR.C.4Rational and Radical EquationsA-REI.A.2

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