2-03 Graph Quadratic Inequalities (3.6)

Example 1

Graph \(y>-\frac{1}{4}(x+1)(x-3)\).

Solution

This is in intercept form, y = a(x – p)(x – q) with \(a=-\frac{1}{4}\) so it opens down and is wider. The x-intercepts are p and q which are −1 and 3.

Graph the inequality as if it was a function.

Find the vertex. For intercept form, the x-coordinate of the vertex is \(x=\frac{p+q}{2}\).

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

Make a table of value around the vertex.

x	−4	−3	−2	−1	0	1	2	3	4	5	6
y	−5.25	−3	−1.25	0	0.75	1	0.75	0	−1.25	−3	−5.25

Graph the points and draw a parabola through the points.

Decide whether the line is solid or dashed. This inequality is > which does not include equality, so the line is dashed.

Decide where to shade.

Method 1: Pick a point not on the line such as (0, 0) and substitute it into the inequality.

$$ y>-\frac{1}{4}\left(x+1\right)\left(x-3\right) $$

$$ 0>-\frac{1}{4}\left(0+1\right)\left(0-3\right) $$

$$ 0>\frac{3}{4} $$

This is false, so shade the other side of the parabola from the point (0, 0).

OR Method 2: The inequality is already solved for y. It is y >, so shade above the parabola.

Example 2

Graph y ≥ (x − 1)² − 2.

Solution

This is in standard form, y = a(x − h)² + k with a = 1 so it opens up.

Graph the inequality as if it was a function.

Find the vertex. For standard form, the vertex is (h, k) = (1, −2).

Make a table of value around the vertex.

x	−2	−1	0	1	2	3	4
y	7	2	−1	−2	−1	2	7

Graph the points and draw a parabola through the points.

Decide whether the line is solid or dashed. This inequality is ≥ which does include equality, so the line is solid.

Decide where to shade.

Method 1: Pick a point not on the line such as (0, 0) and substitute it into the inequality.

y ≥ (x − 1)² − 2

0 ≥ (0 − 1)² − 2

0 ≥ −1

This is true, so shade the side of the parabola containing the point (0, 0).

OR Method 2: The inequality is already solved for y. It is y ≥, so shade above the parabola.

Example 3

Solve the system \(\left\{\begin{align} y& > x^2-4x \\ y&≤-x^2-2x+3 \end{align}\right.\).

Solution

Start by graphing the first inequality, y > x² − 4x.

This is in general form with a = 1, b = −4, and c = 0. The x-coordinate of the vertex is \(x=-\frac{b}{2a}\).

$$ x=-\frac{-4}{2\left(1\right)}=2 $$

The table of values is

x	−1	0	1	2	3	4	5
y	5	0	−3	−4	−3	0	5

The inequality is >, so the line is dashed. And, the inequality is solved for y and is in the form y >, so shade above the parabola.

Now graph the second inequality, y ≤ −x² − 2x + 3, on the same graph. This is also in general form with a = −1, b = −2, and c = 3. The x-coordinate of the vertex is \(x=-\frac{b}{2a}\).

$$ x=-\frac{-2}{2\left(-1\right)}=-1 $$

The table of values is

x	−4	−3	−2	−1	0	1	2
y	−5	0	3	4	3	0	−5

The inequality is ≤, so the line is solid. And, the inequality is solved for y and is in the form y <, so shade below the parabola.

The solution is only the overlap of the two shaded areas.