Question: Sketch $y = (x-2)^2 - 4$, stating the turning point and the intercepts.

Step 1. The form $y = a(x-h)^2 + k$ hands you the turning point with no work:

$a = 1 > 0$, so the parabola opens upwards (∪) and this is a minimum.

Step 2. $x$-intercepts — sub $y = 0$ and keep the square intact:

Step 3. $y$-intercept — sub $x = 0$:

(The curve passes through the origin — the $x$- and $y$-intercept coincide here.)

Step 4. Sketch: ∪-shape with vertex $(2, -4)$, cutting the axis at $0$ and $4$; the line of symmetry is $x = 2$, midway between the roots — a built-in consistency check. ✓

⚠ Watch out: In $(x-2)^2 - 4$ the vertex is $(+2, -4)$ — the $x$-coordinate takes the opposite sign to the one inside the bracket, the constant keeps its sign. And when solving, square-root with $\pm$; expanding the bracket first is slower and invites sign errors.

Question: Sketch $y = -2(x-4)(x+1)$, stating the intercepts and the turning point.

Step 1. $x$-intercepts — read them straight off the factors:

Step 2. $y$-intercept — substitute $x = 0$:

Step 3. Turning point — by symmetry it sits midway between the roots:

Turning point $\left(\dfrac{3}{2},\ 12\dfrac{1}{2}\right)$.

Step 4. Shape: the $-2$ in front makes the parabola open downwards (∩), so this turning point is a maximum. Sketch through $(-1,0)$, $(0,8)$, peak at $(1.5, 12.5)$, down through $(4,0)$.

⚠ Watch out: The factor $(x-4)$ gives the root $x = +4$ — the sign flips. And check the leading coefficient before drawing: a negative out front means ∩, and the midpoint-of-roots trick for the turning point works only because a parabola is symmetric.