Quadratics: Zeros and the Quadratic Formula

Quadratic functions are the first functions in this course whose rate of change is not constant. Before studying that in Topic 1.3, you need four algebra skills: solving with factors, solving with the quadratic formula, finding the vertex, and completing the square.

Quadratics and parabolas

A quadratic function has the form $f(x) = ax^2 + bx + c$ with $a \ne 0$. Its graph is a parabola: a U shape that opens up when $a > 0$ and down when $a < 0$. The turning point is the vertex, and the parabola is symmetric about the vertical line through it, the axis of symmetry.

Solving with factors

If a product is $0$, one of its factors is $0$. So when $ax^2 + bx + c$ can be written as a product, each factor gives a solution. For $x^2 - 5x + 6$, look for two numbers with product $6$ and sum $-5$: $-2$ and $-3$. Then $(x - 2)(x - 3) = 0$, so $x = 2$ or $x = 3$.

The equation must equal $0$ first. To solve $x^2 = 5x - 6$, move everything to one side before you look for factors.

The quadratic formula and the discriminant

Every equation $ax^2 + bx + c = 0$ can be solved with the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The number under the square root, $b^2 - 4ac$, is the discriminant. It tells you how many real solutions there are without solving:

The vertex and completing the square

The vertex of $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. For $y = 2x^2 - 8x + 5$: $x = -\frac{-8}{4} = 2$, and $y = 8 - 16 + 5 = -3$, so the vertex is $(2, -3)$.

Completing the square rewrites a quadratic in vertex form $a(x - h)^2 + k$, where the vertex $(h, k)$ can be read off.

Vertex form also makes solving easy: $(x + 3)^2 - 4 = 0$ gives $(x + 3)^2 = 4$, so $x + 3 = \pm 2$ and $x = -1$ or $x = -5$. Remember the $\pm$; a square root has two signs.

To build a parabola from its vertex and one more point, start with $y = a(x - h)^2 + k$ and use the point to find $a$. With vertex $(-1, 2)$ and point $(0, 5)$: $5 = a(1)^2 + 2$, so $a = 3$ and $y = 3(x + 1)^2 + 2$.