Intersection of a Line and a Quadratic Curve

Line-meets-curve is solved by setting them equal to eliminate $y$, giving a quadratic in $x$. Solve it, then read off each point’s $y$ from the line. Two real roots mean two intersection points (a chord); a repeated root would mean a tangent.

Find the coordinates of the points where the line $y = x + 4$ meets the curve $y = x^2 - 2x$. [5]

The working

Step 1, set line equal to curve: $x^2 - 2x = x + 4 \quad \text{(M1)}$

Step 2, rearrange to a quadratic $= 0$: $x^2 - 2x - x - 4 = 0 \;\Rightarrow\; x^2 - 3x - 4 = 0 \quad \text{(A1)}$

Step 3, factorise and solve: $(x - 4)(x + 1) = 0 \;\Rightarrow\; x = 4 \text{ or } x = -1 \quad \text{(M1)}$

Step 4, find each $y$ from the line $y = x + 4$: $x = 4 \Rightarrow y = 8, \qquad x = -1 \Rightarrow y = 3$

The line meets the curve at $(4, 8)$ and $(-1, 3)$. (A1, A1)

Where the marks are won and lost

• Collect to $= 0$ correctly: $x^2 - 2x - x - 4 = x^2 - 3x - 4$. A sign slip on the $-x$ or $-4$ breaks the quadratic.
• Find each $y$ from the line, not the curve, it’s simpler and avoids mispairing coordinates.
• Give both coordinate pairs; $x$-values alone are incomplete.

Common mistakes

• Sign errors moving $x + 4$ across (the middle term should be $-3x$).
• Finding $x$ but forgetting the $y$-coordinates.
• Substituting into the curve and choosing the wrong $y$.

Full method: Line–Curve Intersection notes. See also Points of Intersection. Topic home: Quadratic Functions pillar.