Lesson 30: Applying Transformations to the Parabola and Determining the Minimum or Maximum

New Concept

The vertex form of a quadratic function is $f(x) = a(x - h)^2 + k$.

Why it matters

This vertex form isn't just about parabolas; it's your first look at a universal principle of transforming any function by adjusting simple parameters. Mastering this unlocks the ability to model and optimize real-world scenarios, from the path of a projectile to the profits of a business.

What’s next

Next, you’ll use this form to instantly identify a parabola's vertex, sketch its graph, and find its minimum or maximum value.

A transformation is a change in the location or shape of a graph. One type of transformation is a shift, or slide. In $f(x) = a(x - h)^2 + k$, the value of $h$ indicates a horizontal shift and $k$ indicates a vertical shift.

The graph of $y = (x - 7)^2$ is shifted 7 units to the right from the parent function $y = x^2$.
The graph of $y = x^2 + 5$ is the parent function $y = x^2$ shifted 5 units up.
The graph of $y = (x + 2)^2 - 3$ is shifted 2 units to the left and 3 units down.

Imagine your parabola is a video game character at the origin. The 'h' in $(x-h)^2$ moves it left or right, but watch out—it's opposite of the sign you see! The 'k' value is simple, moving it up for positive and down for negative. This is how we position the parabola's vertex anywhere on the graph.

The vertex form of a quadratic function is $f(x) = a(x - h)^2 + k$ where $a \neq 0$, and $(h, k)$ represents the vertex of the parabola. The value of $h$ indicates a horizontal shift and $k$ indicates a vertical shift.

In $f(x) = (x - 5)^2 + 2$, the vertex is at $(5, 2)$, representing a shift 5 units right and 2 units up.
For the function $g(x) = -2(x + 1)^2 - 4$, the vertex is at $(-1, -4)$.
For $y = x^2 + 6$, which is $y = (x - 0)^2 + 6$, the vertex is at $(0, 6)$, a vertical shift of 6 units.

This handy formula is a cheat code for graphing parabolas! Instead of plotting tons of points, you can instantly find the vertex at $(h, k)$. The 'h' value slides the graph left or right, and 'k' moves it up or down. It's the ultimate shortcut to finding the parabola's home base on the coordinate plane.

When $a$ is positive in $f(x) = a(x-h)^2+k$, the vertex is a low point and its $y$-coordinate is the minimum value of the function. When $a$ is negative, the vertex is a high point and its $y$-coordinate is the maximum value of the function.

In $f(x) = 2(x - 3)^2 + 5$, since $a=2$ is positive, the parabola opens up and has a minimum value of 5.
In $g(x) = -0.5(x + 1)^2 + 10$, since $a=-0.5$ is negative, it opens down and has a maximum value of 10.
A stock's value is modeled by $V(t) = -5(t-4)^2 + 150$. The stock has a maximum value of 150 dollars.

Think of the parabola's mood! If 'a' is positive, the parabola opens upward like a smile, so its vertex is the lowest point (a minimum). If 'a' is negative, it opens downward like a frown, making the vertex the highest point (a maximum). The sign of 'a' tells you if you're looking for a valley or a peak.

When $0 < |a| < 1$, the parabola is vertically compressed, so it is wider. When $|a| > 1$, the parabola is vertically stretched, so it is narrower.

The graph of $y = 4x^2$ (where $|a|=4 > 1$) is narrower, or vertically stretched, compared to $y = x^2$.
The graph of $y = \frac{1}{4}x^2$ (where $|a|=\frac{1}{4} < 1$) is wider, or vertically compressed, compared to $y = x^2$.
Comparing $y = -5x^2$ and $y = -0.2x^2$, the graph of $y = -5x^2$ is narrower because $|-5| > |-0.2|$.

The absolute value of 'a' is the parabola's personal trainer! If $|a|$ is greater than 1, the parabola is tall and skinny (stretched). If $|a|$ is a fraction between 0 and 1, the parabola is short and wide (compressed). It's all about how intense the curve gets, and 'a' is the intensity dial.