A line and a parabola can meet in two points, one point, or not at all. Finding the meeting points combines systems of equations with solving quadratics.
Why it works
At a meeting point, the same $(x, y)$ satisfies both equations. If both are solved for $y$, the two right sides must be equal there: $x^2 - 2x + 1 = x + 5$. That is a quadratic equation in $x$ alone (substitution). Its solutions are the $x$-coordinates of the meeting points, and each one gives a $y$ from either equation. The number of real solutions, two, one or none, is the number of meeting points.
Key idea
Set the two expressions for $y$ equal, solve the quadratic for $x$, then substitute each $x$ back (into the line, which is simpler) to get its $y$. Report points.
Worked examples
Worked example: Two meeting points
Solve the system $y = x^2 - 2x + 1$ and $y = x + 5$.
- Set equal: $x^2 - 2x + 1 = x + 5$, so $x^2 - 3x - 4 = 0$.
- Factor: $(x - 4)(x + 1) = 0$, so $x = 4$ or $x = -1$.
- From the line: $y = 9$ and $y = 4$.
Answer: $(-1, 4)$ and $(4, 9)$
Worked example: Tangent line
Where does $y = 2x$ meet $y = x^2 + 1$?
- $x^2 + 1 = 2x$, so $x^2 - 2x + 1 = 0$, which is $(x - 1)^2 = 0$.
- One solution, $x = 1$, so the line just touches the parabola at $(1, 2)$.
Answer: One point, $(1, 2)$.
Worked example: No meeting point
Does the line $y = x$ meet the parabola $y = x^2 + 3$?
- Set equal: $x^2 + 3 = x$, so $x^2 - x + 3 = 0$.
- Discriminant: $(-1)^2 - 4(1)(3) = 1 - 12 = -11 < 0$, so there is no real solution.
- The parabola stays above the line everywhere.
Answer: No: the system has no solution.
Common mistake
Stopping at the $x$-values. The system asks for points, so each $x$ needs its $y$. Check each point in both original equations.