Every quadratic is the same curve, moved.

Here is $y = x^2 - 4x - 5$. Its shape is fixed — a parabola. Everything else about it, where it turns and where it crosses, is position, and the algebra tells you the position.

scroll ↓

$a = 1$ is positive, so it opens upwards. Factorise to $(x-5)(x+1)$ and the roots are $5$ and $-1$; the constant gives the $y$-intercept $-5$. The axis of symmetry sits midway between the roots, at $x = 2$, and the vertex is on it.

Completing the square rewrites the same curve with its vertex on show: $(x-2)^2 - 9$ says $(2,\,-9)$. A square is never negative, so $-9$ is the minimum.

Here $\Delta = 16 - 4c$. As $c$ rises the roots close in, until at $c = 4$ they merge: $\Delta = 0$, one repeated root, the curve just touches the axis.

Read it as a picture. $-\tfrac{b}{2a}$ is the axis of symmetry; $\pm\tfrac{\sqrt{\Delta}}{2a}$ is how far each root sits either side of it. Here: $2 \pm 3$. And when $\Delta$ is negative there's no square root to take — no roots, exactly as the picture showed.

One shape. Three numbers place it.

Factorising with a leading coefficient, quadratic inequalities, disguised quadratics, and twelve graded exam questions with full worked solutions are waiting.

Read the full chapter › Jump to the questions ›