Solve Quadratic Inequalities Graphically
This is a tutorial on how to solve quadratic inequalities graphically. The quadratic inequalities explored are of the type
\[ a x^2 + b x + c \lt 0 \]
and
\[ a x^2 + b x + c \gt 0 \]
Review
An app. plots the graph of \( y = a x^2 + b x + c \) and displays part of the graph that is below the x axis \( ( y \lt 0 ) \) in blue and part of the graph that is above the x axis \( ( y \gt 0 ) \) in red. To solve a quadratic inequality you just read the interval corresponding to \( y \lt 0 \) or \( y \gt 0 \) depending on the inequality to solve.
Interactive Tutorials
Quadratic Function Plotter
Enter values for coefficients a, b, and c to see how they affect the parabola shape and position.
$$y = ax^2 + bx + c$$
Current equation: \( y = 1.0 x^2 + 2 x -3 \)
Example 1 :
Solve graphically and analytically the quadratic inequality
\[
- x^2 + 3x + 4 \lt 0
\]
Solution to Example 1:
Graphical solution:
Use the app. above to enter the coefficients \( a = -1 \), \( b = 3 \) and \( c = 4 \) and graph the equation \( y = - x^2 + 3x + 4 \). The solution set to the inequalit \( - x^2 + 3x + 4 \lt 0 \) correspond to the x coordinates of the points on the graph for which \( y \lt 0 \) BLUE. We have two intervals for which \( y \lt 0 \) whose union is written in interval form as:
\[
(-\infty , -1) \cup (4 , +\infty)
\]
Analytical solution:
Factor the left hand term of the given inequality
\[
- x^2 + 3x + 4 = (x + 1)(-x + 4)
\]
To solve the given inequality, we study the sign of the expression:
\[
(x + 1)(-x + 4)
\]
This expression has zeros at \(x = -1\) and \(x = 4\). These values divide the number line into three intervals:
- \((-\infty, -1)\)
- \((-1, 4)\)
- \((4, +\infty)\)
We analyze the sign of the expression \(-x^2 + 3x + 4\) on each interval by choosing test points:
1. Interval \((-\infty, -1)\):
Let \(x = -2\)
\[
-x^2 + 3x + 4 = -(-2)^2 + 3(-2) + 4 = -4 - 6 + 4 = -6
\]
The result is negative, so this interval is part of the solution set.
2. Interval \((-1, 4)\):
Let \(x = 0\)
\[
-x^2 + 3(0) + 4 = 0 + 0 + 4 = 4
\]
The result is positive, so this interval is not part of the solution set.
3. Interval \((4, +\infty)\):
Let \(x = 5\)
\[
-x^2 + 3(5) + 4 = -25 + 15 + 4 = -6
\]
The result is negative, so this interval is part of the solution set.
Final Answer:
The solution set of the inequality, where the expression is negative, is the union of the two intervals over which the expression \( - x^2 + 3x + 4 \) is negative :
\[
(-\infty, -1) \cup (4, +\infty)
\]
- Both the graphical and analytical methods give the same answer.
Example 2
Solve graphically and analytically the quadratic inequality
\[ -x^2 + 4x - 5 \gt 0 \]
Solution to Example 2:
Graphical solution:
Use the app. to set coefficients \( a = -1\), \( b = 4 \) and \( c = -5 \) and graph the equation \( y = -x^2 + 4x - 5 \). This inequality has no solutions since the whole graph is below the x axis and therefore \( -x^2 + 4x - 5 \lt 0 \) for all values of \( x \).
Analytical solution
To solve the given inequality, we analyze the sign of the quadratic expression.
-
The expression \(-x^2 + 4x - 5\) cannot be factored over the real numbers because its discrminant \[ \Delta = 4^2 - 4 (-1)(-5) = -4 \] is negative.
Therefore, it has no real zeros and its sign does not change across the real number line.
To determine its sign, we evaluate the expression at a single value of \(x\).
Let's choose \(x = 0\):
\[
-x^2 + 4x - 5 = -(0)^2 + 4(0) - 5 = -5
\]
Since the result is negative, the expression is negative for all real values of \(x\).
-
The expression \(-x^2 + 4x - 5\) is always negative. Therefore, the inequality
\(-x^2 + 4x - 5 > 0\) has no solution.
Exercises
Solve each quadratic inequality both graphically (using the app) and analytically:
- \[
-x^2 - 4x \lt -5
\]
- \[
x^2 - 2x + 8 \geq 0
\]
- \[
x^2 - 3x \leq 0
\]
Solutions to the Above Exercises
- \[
(-\infty , -5) \cup (1 , +\infty)
\]
- \[
(-\infty , +\infty)
\]
- \[
[0 , 3]
\]
More References and Links