Quadratic functions: factoring and the graph

Sketch parabolas from factored form: zeroes, axis of symmetry, vertex, concavity and y-intercept, plus double and distinct roots.

The parabola and its features

A quadratic function has the form $y = ax^2 + bx + c$ with $a \ne 0$. Its graph is a parabola. Every parabola has the same five features, and a full sketch labels all of them:

Concavity: if $a > 0$, concave down if $a < 0$.
$y$-intercept: put $x = 0$, which gives $y = c$.
Zeroes ($x$-intercepts): where $y = 0$.
Axis of symmetry: the vertical line through the middle of the parabola.
Vertex: the turning point, sitting on the axis of symmetry.

Factored form shows the zeroes

If a quadratic factors as $y = a(x - \alpha)(x - \beta)$, its zeroes are $x = \alpha$ and $x = \beta$, because a product is zero only when a factor is zero.

Build a parabola from its zeroes

Try this

Move α and β. Check the axis of symmetry is always halfway between them.
Make α equal to β. What happens to the graph at that zero?
Make $a$ negative. What changes, and what stays the same?

Example 1●

Sketch $y = x^2 - 2x - 8$, showing all intercepts and the vertex.

Factor: $y = (x - 4)(x + 2)$.

Find two numbers that multiply to −8 and add to −2: they are −4 and 2.

Zeroes: $x = 4$ and $x = -2$.

Each factor gives a zero.

$y$-intercept: $y = -8$.

Put x = 0.

Axis of symmetry: $x = \dfrac{-2 + 4}{2} = 1$.

Halfway between −2 and 4.

Vertex: $y = 1 - 2 - 8 = -9$, so the vertex is $(1, -9)$.

Substitute x = 1 into the original function.

Concave up. Sketch with $(-2, 0)$, $(4, 0)$, $(0, -8)$ and $(1, -9)$ labelled.

a = 1 > 0. Plot all four labelled points, then draw a smooth U shape through them.

Check your sketch

This is the finished sketch of Example 1. Hover over it to read off other points.

Example 2●●

Sketch $y = -2x^2 + 4x + 6$.

$y = -2(x^2 - 2x - 3) = -2(x - 3)(x + 1)$

Take out the common factor −2 first. This also makes the bracket easier to factor.

Zeroes $x = 3$ and $x = -1$. $y$-intercept $6$.

Axis $x = \dfrac{3 + (-1)}{2} = 1$. Vertex: $y = -2 + 4 + 6 = 8$, so $(1, 8)$.

Concave down, with maximum value $8$.

a = −2 is negative, so the parabola opens downward and the vertex is a maximum.

Double zeroes and perfect squares

When both factors are the same, as in $y = (x - 3)^2$, there is only one zero. We call it a . The parabola touches the $x$-axis at its vertex instead of crossing it.

When the two zeroes are different, we call them distinct. A quadratic always has two distinct zeroes, one double zero, or no zeroes at all.

Finding a quadratic from its zeroes

If the zeroes are $\alpha$ and $\beta$, the quadratic is $y = a(x - \alpha)(x - \beta)$ for some $a \ne 0$. One more point on the graph fixes $a$. The range runs from the vertex: $y \ge$ (vertex $y$) if concave up, $y \le$ (vertex $y$) if concave down.

Example 3●●

A parabola has zeroes $1$ and $5$ and $y$-intercept $10$. Find its equation and range.

$y = a(x - 1)(x - 5)$

$10 = a(-1)(-5) = 5a$, so $a = 2$ and $y = 2(x - 1)(x - 5)$.

Substitute the point (0, 10).

Vertex $(3, -8)$, concave up, so the range is $y \ge -8$.

Axis x = 3, so the vertex y-value is 2(2)(−2).

Flashcards

Zeroes vs roots

Functions have zeroes; equations have roots.

Axis of symmetry from zeroes α and β

$x = \dfrac{\alpha + \beta}{2}$

Concave up or down?

Up if $a > 0$ (cup), down if $a < 0$ (cap).

y-intercept of y = ax² + bx + c

$c$

Double zero

From a repeated factor like $(x - k)^2$. The graph touches the axis at $x = k$.

Practice

Find the zeroes of $y = x^2 - 4x - 5$. Enter both, separated by a comma.
The zeroes of a parabola are $x = -3$ and $x = 7$. Its axis of symmetry is:
What is the vertex of $y = (x - 2)(x - 6)$?
The graph of $y = -(x + 1)^2$:
The $y$-intercept of $y = (x - 3)(x + 5)$ is:
Find the axis of symmetry of $y = x^2 - 6x + 5$ from its zeroes.
A parabola has zeroes $2$ and $6$ and passes through $(0, 12)$. Its leading coefficient is:
The range of $y = (x - 1)(x - 5)$ is: