Finding the Low Point: A Complete Guide to Minimums on a Graph
What is a Minimum, Visually?
Imagine walking along a smooth, curved path over a hill. You would walk down into a valley and then back up the other side. The lowest point in that valley, the very bottom, is what mathematicians call a minimum. On a graph, it is a point where the curve changes direction from going downward to going upward. This makes it a type of turning point. The key feature is that if you look at the point just to its left and the point just to its right, both are higher up on the graph. That means the $y$-value (the vertical coordinate) at the minimum is the smallest value in that immediate neighborhood.
The simplest graph with a clear minimum is a parabola that opens upward. Its equation is often in the form $y = ax^2 + bx + c$, where $a > 0$. The lowest point on this curve is called the vertex. For example, the graph of $y = x^2 - 4x + 3$ has its vertex at $(2, -1)$. You can see that at $x = 2$, the $y$-value is $-1$. For $x = 1.9$ and $x = 2.1$, the $y$-values are slightly higher, confirming $(2, -1)$ is a minimum.
The Language of Minima: Local vs. Global
It's important to distinguish between two types of minimums: local (or relative) and global (or absolute).
| Type of Minimum | Definition | Simple Example |
|---|---|---|
| Local Minimum | A point whose $y$-value is lower than all other points nearby. It is the lowest point in its immediate neighborhood. | A small dip or valley between two hills on a roller coaster track. It's low compared to the track right before and after it, but not necessarily the lowest point on the whole ride. |
| Global Minimum | A point whose $y$-value is lower than all other points on the entire graph within the domain[1] we are considering. | The absolute deepest valley on a map of a mountain range. No other point on the map is lower. |
A graph can have multiple local minima, but only one global minimum (or none, if the graph goes down forever). The global minimum is also a local minimum, but a local minimum is not always the global one.
Using Slope and Calculus to Find the Minimum
For smooth, continuous curves, we can use the concept of slope to find minimum points precisely. The slope tells us how steep the curve is at any point.
- Slope Before the Minimum: As you approach a minimum from the left, the curve is going downhill. This means the slope is negative ($< 0$).
- Slope at the Minimum: At the very bottom of the valley, the curve is momentarily flat. This is the turning point. The slope is exactly zero ($= 0$).
- Slope After the Minimum: As you leave the minimum to the right, the curve starts going uphill. This means the slope becomes positive ($> 0$).
This pattern of slope (negative $\to$ zero $\to$ positive) is the key signature of a minimum.
1. Find where the derivative equals zero: Solve $f'(x) = 0$. These are called critical points[2].
2. Test the sign of the derivative just before and after each critical point. If the sign changes from negative to positive, that point is a local minimum.
Example: Let's find the minimum of $f(x) = x^2 - 6x + 8$.
1. Find the derivative: $f'(x) = 2x - 6$.
2. Set derivative to zero: $2x - 6 = 0$, so $x = 3$.
3. Test the slope pattern:
• For $x = 2.9$, $f'(2.9) = 2(2.9) - 6 = -0.2$ (negative).
• For $x = 3.1$, $f'(3.1) = 2(3.1) - 6 = 0.2$ (positive).
Slope changes from negative to zero to positive, so $x = 3$ gives a minimum. The point is $(3, f(3)) = (3, -1)$.
A Practical Application: Minimizing Cost or Maximizing Efficiency
Finding minimums isn't just a math exercise; it's a powerful tool for solving real-world problems. A classic application is in business and engineering: minimizing cost or maximizing efficiency.
Scenario: Imagine a farmer needs to build a rectangular pen for animals using a fixed length of fencing, say $100$ meters. She wants to enclose the maximum possible area. This is a "min-max" problem—we want to maximize area, but we can approach it by thinking about the minimum of the negative area function, or by direct analysis.
Let the width be $w$ and the length be $l$. The perimeter is $2w + 2l = 100$, so $l = 50 - w$.
The area $A$ is: $A(w) = w \times l = w(50 - w) = 50w - w^2$.
This is a quadratic function ($a = -1 < 0$), so its graph is an upside-down parabola. The maximum area is at the vertex, which is also the point where the slope is zero.
1. Find derivative: $A'(w) = 50 - 2w$.
2. Set to zero: $50 - 2w = 0$ gives $w = 25$ meters.
3. Check slope: For $w=24$, $A'(24)=2$ (positive). For $w=26$, $A'(26)=-2$ (negative). Slope changes from positive to zero to negative, confirming this is a maximum point. The maximum area is $A(25) = 625$ square meters.
This shows that the most efficient shape (giving max area for fixed perimeter) is a square. By finding where the slope of the area function is zero, we solved a practical optimization problem.
Important Questions About Minimums
Q1: Can a minimum occur at a sharp corner or cusp on a graph?
Q2: How is finding a minimum related to solving real problems like saving fuel or time?
Q3: What is the difference between a minimum and just a low point on a graph?
Conclusion
Footnote
[1] Domain: The set of all possible input values ($x$-values) for a function. When searching for a global minimum, we only consider points within this defined domain.
[2] Critical Point: A point on the graph of a function where its derivative is either zero or undefined. These are the candidates for local maxima and minima.