Algebra · Graphs of functions

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Quadratic graphs

Why a quadratic draws a symmetric parabola, what the roots, y-intercept and turning point each are on the picture, and how the graph solves equations by intersection.

Algebra · Graphs of functions

Quadratic graphs

Why a quadratic draws a symmetric parabola, what the roots, y-intercept and turning point each are on the picture, and how the graph solves equations by intersection.

Why it works

Plot

y = x^2 - 2x - 3

by feeding in

x

-values:

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$5$	$0$	$-3$	$-4$	$-3$	$0$	$5$

The values fall then rise, and they MIRROR —

5, 0, -3, -4, -3, 0, 5

. That symmetry isn't luck: squaring treats

+

and

-

alike, so every parabola has a vertical line of symmetry through its lowest (or highest) point. The curve is smooth and U-shaped — never pointed, never made of line segments.

The picture is a dictionary of the algebra:

Roots — where the curve crosses the $x$ -axis ( $y = 0$ ): here

x = -1

and

x = 3

, exactly the solutions of

x^2 - 2x - 3 = 0

$y$ -intercept — where it crosses the $y$ -axis ( $x = 0$ ): the

constant term,

(0, -3)

Turning point — the bottom of the U. By symmetry its $x$ -value is

MIDWAY between the roots:

x = \frac{-1 + 3}{2} = 1

, and

y = 1 - 2 - 3 = -4

: the point

(1, -4)

.

Factorised form hands you the sketch.

y = (x - 2)(x + 4)

: roots where each bracket dies (

2

and

-4

y

-intercept from

x = 0

(

-8

), symmetry line midway at

x = -1

, turning point

(-1, -9)

. Four facts, no table.

Graphs solve equations by intersection. The solutions of

x^2 - 2x - 3 = 2

are the

x

-values where the parabola meets the horizontal line

y = 2

— read them off (about

3.4

and

-1.4

). A graphical answer is an ESTIMATE; the exam says "use the graph", and one decimal place of honesty beats false precision.

A negative $x^2$ -coefficient flips the U upside-down — the turning point becomes a maximum. Everything else reads the same way.