- A non-linear inequality in two variables (usually $x$ and $y$) describes a region of the coordinate plane, not just a single curve.
- The "boundary" of that region is the related equation, and whether the boundary is included depends on the inequality symbol.
- When you solve an inequality such as $$y < x^2 - 2x - 3$$ you are looking for all ordered pairs $(x,y)$ that make the inequality true.
- Graphically:
- The curve $y = x^2 - 2x - 3$ is the boundary.
- Points below the curve satisfy $y < x^2 - 2x - 3$.
- Points above the curve satisfy $y > x^2 - 2x - 3$.
- It helps to read inequalities like "$y$ is less than (the function of $x$)."
- If $y$ must be less, the solution is typically the region below the boundary curve.
A key graphical idea is whether points on the boundary are included.
- If the inequality is strict (for example, $y < f(x)$ or $y > f(x)$), draw the boundary as a dashed curve.
- If the inequality is non-strict (for example, $y \le f(x)$ or $y \ge f(x)$), draw the boundary as a solid curve.
- A very common mistake is drawing a solid boundary for $<$ or $>$, or a dashed boundary for $\le$ or $\ge$.
- The boundary style communicates inclusion.
For many non-linear inequalities, you cannot "solve" them into a single interval easily. The standard approach is graphical.
To solve an inequality like $y < f(x)$ or $y \ge f(x)$:
- Graph the boundary defined by the equality $y=f(x)$.
- Choose a test point not on the curve (often $(0,0)$ if it is not on the boundary).
- Substitute the test point into the original inequality.
- If the inequality is true, shade the region containing the test point.
- If it is false, shade the opposite region.
If you choose $(0,0)$ as a test point, quickly check whether $(0,0)$ lies on the boundary first. If it does, pick a different point such as $(0,1)$ or $(1,0)$.
- The boundary curve splits the plane into separate regions.
- Within any one region, the inequality will be consistently true or consistently false (until you cross the boundary).
- So testing a single point tells you which side to shade.
- Solve by sketching: $$y < -x^2 + 7x - 10$$
- First draw the boundary parabola: $$y = -x^2 + 7x - 10$$
- To sketch accurately, find key features.
- Set $y=0$: $$-x^2 + 7x - 10 = 0$$
- Factorizing: $$-(x^2 - 7x + 10) = -(x-2)(x-5)$$
- So the $x$-intercepts are $x=2$ and $x=5$.
- For $y=ax^2+bx+c$, the $x$-coordinate of the vertex is $x=-\frac{b}{2a}$.
- Here $a=-1$ and $b=7$, so $$x=-\frac{7}{2(-1)}=3.5$$
- Substitute to get $y$: $$y = -(3.5)^2 + 7(3.5) - 10 = -12.25 + 24.5 - 10 = 2.25$$
- So the vertex is $(3.5, 2.25)$.
- Because the inequality is strict ($<$), draw the parabola dashed.
- Pick a test point not on the curve, for example $(3,0)$: $$-x^2+7x-10 = -(3)^2 + 7(3) - 10 = -9+21-10=2$$
- The inequality checks as: $$0 < 2$$ which is true.
- Therefore, shade the region containing $(3,0)$, which corresponds to the region below the parabola.
- For quadratics, quickly sketch using: (1) opening direction from the sign of $a$, (2) roots if available, and (3) vertex.
- A better sketch makes it much easier to shade the correct region.
The same graph-test-shade idea works for other non-linear boundaries.
- For example: $$y > \ln(x-2) + 3$$
- The boundary $y=\ln(x-2)+3$ only exists for $x-2>0$, so $x>2$.
- When the boundary involves $\ln(x-2)$ (or any logarithm), always state and respect the domain.
- You cannot shade solutions where the expression is undefined.
- For example: $$y \le e^{x+1} - 3$$
- Draw the exponential curve $y=e^{x+1}-3$ as a solid boundary (because of $\le$), then shade the region on or below it.
- For example: $$y < \frac{x+3}{x-3}$$
- The boundary is undefined at $x=3$, giving a vertical asymptote at $x=3$.
- Since the function is undefined there, the vertical asymptote is always drawn as a dashed line, even if the inequality is non-strict ($\le$ or $\ge$).
- The curve will typically have two branches.
- With rational inequalities, the vertical asymptote splits the plane into separate regions.
- To determine the correct shading, test a point on each side of the vertical asymptote (e.g., one point where $x < 3$ and one where $x > 3$) to verify the solution set across the entire domain.
A system of inequalities means multiple inequalities must be true at the same time. Graphically, you:
- Graph and shade the solution region for each inequality.
- The solution to the system is the intersection (overlap) of the shaded regions.
To solve graphically: $$y < x+2, \quad y \ge 3-2x$$
- $y=x+2$ is dashed (strict inequality).
- $y=3-2x$ is solid (non-strict inequality).
- Shade below the dashed line and on/above the solid line, then take the overlap.
- Even though this example is linear, the "overlap idea" is exactly how you handle systems where one inequality is non-linear (for example, a line together with a parabola).
- You shade each, then keep only the common region.
Sometimes you are given a shaded graph and asked to write an inequality.
A good approach:
- Identify the boundary curve (parabola, line, exponential, etc.). Write its equation (often by using intercepts, vertex, or key points).
- Decide whether the boundary is included (solid) or excluded (dashed).
- Determine if the region is above/below (or left/right) of the boundary by checking one point in the shaded region.
- If a parabola is dashed and the shading is below it, the inequality will have the form $y<f(x)$.
- If it is solid with shading above, it is $y\ge f(x)$.
- For $y < f(x)$, do you automatically think "below," but still confirm with a test point?
- Can you explain what dashed vs solid means in one sentence?
- Can you solve a system by shading each inequality and identifying the overlap?
- If given a shaded region, can you write an inequality and justify it using a test point?