Skill 11 of 12

Key features of quadratic functions and graphs

Identify the vertex, axis of symmetry, maximum or minimum, intercepts, and direction of opening of a quadratic function from its equation in standard, vertex, or factored form, and from its graph.

Common Core standards: F-IF.4F-IF.7aF-IF.8a
Builds on: Solve quadratic equations by factoring
Complete the square to write vertex form

Common mistakes

What students most often get wrong with this skill:

Reading the vertex of $(x-2)^2+3$ as $(-2,3)$
Confusing x-intercepts with the y-intercept
Saying the parabola opens downward when $a$ is positive
Using $-b/a$ instead of $-b/(2a)$ for the axis of symmetry

A worked example

From the lesson, with every step explained.

Example 1: Vertex form

Problem: For $y=(x+1)^2-4$ , find the vertex, axis of symmetry, direction of opening, minimum or maximum, and intercepts.

Plan: Match the equation to $y=a(x-h)^2+k$ . Then use the intercept rules.

1) Rewrite $x+1$ as $x-(-1)$ .
This shows $h=-1$ . The sign flips, so $h$ is not $1$ .
2) Here $k=-4$ and $a=1$ .
So the vertex is $(-1,-4)$ .
3) The axis of symmetry is $x=-1$ .
4) Since $a=1>0$ , the parabola opens up.
So $-4$ is the minimum value.
5) For the y-intercept, set $x=0$ .

y=(0+1)^2-4=1-4=-3

The y-intercept is $(0,-3)$ .
6) For the x-intercepts, set $y=0$ .

(x+1)^2=4

Take the square root of both sides.

x+1=2 \text{ or } x+1=-2

So $x=1$ or $x=-3$ .

Answer: vertex $(-1,-4)$ , axis $x=-1$ , opens up, minimum $-4$ , x-intercepts $1$ and $-3$ , y-intercept $-3$ .

Check: Substitute $x=1$ . Then $(1+1)^2-4=4-4=0$ .
Substitute $x=-3$ . Then $(-3+1)^2-4=4-4=0$ . Both work.

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