Sketching a Parabola: Intercepts and Vertex

To sketch a quadratic you need its key features: the $x$-intercepts, the $y$-intercept, and the vertex. Factorising gives the roots, the constant gives the $y$-intercept, and completing the square gives the vertex.

The curve $y = x^2 - 2x - 8$. Find its $x$-intercepts, $y$-intercept and the coordinates of its vertex. [5]

The working

Step 1, $x$-intercepts (set $y = 0$ and factorise): $x^2 - 2x - 8 = (x - 4)(x + 2) = 0 \;\Rightarrow\; x = 4, \ x = -2 \quad \text{(M1, A1)}$

Step 2, $y$-intercept (set $x = 0$): $y = 0 - 0 - 8 = -8 \;\Rightarrow\; (0, -8) \quad \text{(A1)}$

Step 3, vertex by completing the square: $x^2 - 2x - 8 = (x - 1)^2 - 1 - 8 = (x - 1)^2 - 9 \;\Rightarrow\; \text{vertex } (1, -9) \quad \text{(M1, A1)}$

The parabola opens upward (positive $x^2$ coefficient), crossing the $x$-axis at $-2$ and $4$, the $y$-axis at $-8$, with its minimum at $(1, -9)$.

Where the marks are won and lost

• The $x$-intercepts come from factorising $y = 0$; the $y$-intercept is the constant term $-8$. Don’t confuse the two.
• The vertex from completing the square: $(x - 1)^2 - 9$ gives $(1, -9)$, note the vertex $x$ is $+1$ (from the $(x-1)$ bracket) and $y = -9$.
• The vertex lies midway between the roots ($x = \frac{-2 + 4}{2} = 1$), a useful check.

Common mistakes

• Swapping the $x$- and $y$-intercepts.
• Reading the vertex $x$-coordinate as $-1$ from $(x - 1)$.
• Sign errors completing the square ($-1 - 8 = -9$).

Full method: Maximum/Minimum & the Vertex notes. Topic home: Quadratic Functions pillar.