Graphing Quadratics

After a basketball leaves your hands, its path is neither a straight line nor a circular arc but a graceful curve that rises, peaks and falls — a parabola. Fountain jets and bridge arches carry its silhouette. Behind the curve stands a very plain function:

$y = ax^2 + bx + c$

Plot the simplest parabola first

Start from $y = x^2$ and make a table of points (drawn on Cartesian coordinates):

$x$	$-2$	$-1$	$0$	$1$	$2$
$y = x^2$	4	1	0	1	4

Connect the dots and a bowl-shaped curve appears. Notice its symmetry: $x = -2$ and $x = 2$ produce the same $y$ — squaring eats the minus sign. The lowest point $(0, 0)$ is the vertex, and the vertical line that splits the curve into mirror halves is the axis of symmetry.

The opening direction reads a

$a > 0$ : the parabola opens upward, like a bowl, and the vertex is the minimum;
$a < 0$ : it opens downward, like an umbrella, and the vertex is the maximum.

$|a|$ controls the build: $y = 2x^2$ already reaches 2 at $x = 1$ , steeper and slimmer than $y = x^2$ , while $y = 0.5x^2$ climbs slowly and looks wider. If $a = 0$ the square term disappears and only $y = bx + c$ remains — the parabola flattens into a straight line. The coefficient $a$ is the soul of the parabola.

Axis of symmetry and vertex

For the general form, the axis of symmetry is the vertical line

$x = -\frac{b}{2a}$

and the vertex sits on it. For example, $y = x^2 - 4x + 3$ has axis $x = -\frac{-4}{2} = 2$ and vertex height $4 - 8 + 3 = -1$ , namely $(2, -1)$ .

Where does the formula come from? From symmetry. Substitute $x = 1$ and you get 0; substitute $x = 3$ and you get 0 again — two equal-height points must sit symmetrically about the axis, and their midpoint $\frac{1+3}{2} = 2$ is exactly it.

Do not drop the minus sign

The leading minus in $x = -\frac{b}{2a}$ is the easiest thing to lose. In $y = x^2 - 4x + 3$ we have $b = -4$ , so $-\frac{-4}{2} = 2$ — a negative times a negative turns positive, and the axis lands on the positive side.

The discriminant counts the intercepts

The number of crossings with the x-axis is settled by the discriminant $b^2 - 4ac$ :

$b^2 - 4ac > 0$ : two intercepts;
$b^2 - 4ac = 0$ : exactly one — the vertex rests on the x-axis;
$b^2 - 4ac < 0$ : no intercepts at all.

For $y = x^2 - 1$ the discriminant is $0 + 4 = 4 > 0$ , with intercepts at $x = \pm 1$ ; for $y = x^2$ it is 0, one intercept; for $y = x^2 + 1$ it is $-4 < 0$ — even the minimum sits 1 above the ground, never reaching the x-axis. An intercept's x-value is a solution of $y = 0$ , which is exactly what quadratic equations are about.

Sketch a whole parabola

Draw $y = x^2 - 2x - 3$ : the axis is $x = \frac{2}{2} = 1$ , the vertex $(1, -4)$ , and the discriminant $4 + 12 = 16 > 0$ promises two intercepts. Plot points: $x = -1$ gives 0, $x = 0$ gives $-3$ , $x = 3$ gives 0. The curve dips from $(-1, 0)$ down to $(1, -4)$ and climbs back to $(3, 0)$ — vertex, axis and intercepts all agree.

When is the ball highest

A thrown ball's height is roughly $h = -5t^2 + 20t$ (meters). Here $a = -5 < 0$ , so it opens downward. The peak sits on the axis of symmetry at $t = \frac{20}{10} = 2$ seconds, where $h = -20 + 40 = 20$ meters. The discriminant $400 > 0$ gives two zeros: launch at $t = 0$ , landing at $t = 4$ . Read off the vertex and both "when" and "how high" are answered at once.

Check yourself

Quick quiz

1. Which way does y = −3x² open?
2. What is the axis of symmetry of y = x² − 6x + 5?
3. How many x-intercepts does y = x² + 2x + 5 have?