| Graphing Inequalities |
If we're asked to solve y = 2x + 3 graphically, we graph the line y = 2x + 3
with a slope of 2 and a y-intercept of 3 and that would be our graphical solution,
since we would have plotted every point in the plane which satisfies the
equation y = 2x + 3. That is: the coordinates (x, y) of every point on that line,
when substituted into the equation would make y = 2x + 3 a true statement.
The points on the line satisfy the equality relation.
So points off the line must satisfy the inequalities.
The line divides the plane into two halves:
points above the line, and the points below the line.
What if we were asked to solve the inequality y > 2x + 3 graphically?
Let's look at the graph of the line y = 2x + 3 to see what we can discover.
Using the coordinates of A(-3, 5) as x and y in the inequality y > 2x + 3,
we get 5 > 2(-3) + 3 or 5 > -3, which is a true statement,
so A(-3, 5) is in the solution region.
If we substitute the coordinates of B(3, 3) into y > 2x + 3,
we find they give the untrue statement 3 > 2(3) + 3 or 3 > 9.
Thus, the solution region for y > 2x + 3 is the shaded area in the graph below.
Note that the line y = 2x + 3 is a dashed line because it is not included in the solution.
Had the question been y ¥ 2x + 3, (including the equality),
we would make the line solid instead of dashed to indicate equality.
Now let's use what we know about graphing parabolas to solve x2 - 3x - 4 > 0.
We'll plot the graph of y = x2 - 3x - 4 then see where the y values are positive.
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Example 1
Solve x2 - 3x - 4 > 0 graphically.
Solution:
y = x2 - 3x - 4 is a parabola opening upwards with zeros (-1, 0) and (4, 0)
y-intercept (0, -4) and axis of symmetry x = 3/2.
The minimum y value is 
, as shown.
The positive y values on the curve occur in the tails; when x < -1 and x > 4.
These, therefore, are the solution intervals.
If we'd been solving x2 - 3x - 4 < 0, the solution would be -1 < x < 4
since that's where the y-values are negative.
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Example 2
Solve - x2 + 2x + 3 > 0 graphically.
Solution:
y = -x2 + 2x + 3 is a parabola opening downwards, with
x-intercepts (-1, 0) and (3, 0), y-intercept (0, 3) and
axis of symmetry the line x = 1 as shown.
We want to find where y is > 0 or positive, so we're looking for x-values
that put the curve above the x-axis. The solution interval is -1 < x < 3.
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Example 3
Solve these inequalities graphically y < -x2 + 2x + 3 and y > x + 1.
As we see, the points of intersection are (2, 3) and (-1, 0).
We want the area below the parabola, above the line.
The x-values that define this region are greater than -1 but less than 2..
Practice
Solve graphically:
1) y > x 2 + 1 and y < 2x + 1.
2) y < x + 3, y ¥ 3 and y < -x + 10
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Solutions
1) the solution interval is: 0 < x < 2
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2) y < x + 3, y ¥ 3 and y < -x + 10
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