When Is a Line a Tangent to a Curve?

This question links three ideas 0606 loves to combine: intersection, the discriminant, and tangency. The unlock is recognising that “tangent” means the line meets the curve exactly once, so the equation you get by setting them equal has equal roots.

The line $y = 2x + k$ is a tangent to the curve $y = x^2 - 4x + 5$. Find the value of the constant $k$. [5]

The working

Step 1, set line equal to curve to find the intersection equation: $x^2 - 4x + 5 = 2x + k \quad \text{(M1)}$

Step 2, rearrange to $= 0$ (all terms on one side): $x^2 - 4x - 2x + 5 - k = 0 \;\Rightarrow\; x^2 - 6x + (5 - k) = 0 \quad \text{(A1)}$

Step 3, apply the tangent condition. One point of contact means equal roots, so $b^2 - 4ac = 0$ with $a = 1$, $b = -6$, $c = 5 - k$: $(-6)^2 - 4(1)(5 - k) = 0 \quad \text{(M1)}$

Step 4, solve for $k$: $36 - 20 + 4k = 0 \;\Rightarrow\; 16 + 4k = 0 \;\Rightarrow\; k = -4 \quad \text{(M1, A1)}$

A quick check: with $k = -4$ the equation is $x^2 - 6x + 9 = 0 = (x - 3)^2$, a genuine repeated root at $x = 3$, confirming a single point of contact.

Where the marks are won and lost

• Collecting to $= 0$ correctly is a mark. The constant term is $5 - k$; a sign slip here (writing $5 + k$) breaks the discriminant.
• "$c = 5 - k$" carries the unknown into the discriminant. Students who set $c = 5$ and ignore $k$ cannot finish.
• The condition is $= 0$ (tangent), not $> 0$ or $< 0$. Decode “tangent” before choosing.

Common mistakes

• Using $b^2 - 4ac > 0$ or $< 0$ instead of $= 0$.
• Sign errors when moving $2x + k$ across (the $x$-coefficient becomes $-6$, the constant $5 - k$).
• Trying to differentiate to find the tangent, valid but slower here, and it still needs the point, which you do not have yet.

Full method: Line–Curve Intersection notes. Topic home: Quadratic Functions pillar.