Quadratic Functions

Hey students! 👋 Ready to dive into one of the most fascinating and practical topics in Algebra 2? Today we're exploring quadratic functions - mathematical powerhouses that help us understand everything from the arc of a basketball shot to maximizing profits in business! By the end of this lesson, you'll be able to graph quadratic functions, transform them using vertex form, complete the square like a pro, and solve real-world problems involving projectile motion and optimization. Let's unlock the secrets of these beautiful parabolic curves! 🚀

Understanding Quadratic Functions and Their Graphs

A quadratic function is any function that can be written in the form $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are real numbers and $a ≠ 0$. The graph of every quadratic function is a U-shaped curve called a parabola! 📈

Think about throwing a basketball toward the hoop - the path the ball follows through the air forms a perfect parabola. This happens because gravity pulls the ball downward at a constant rate while it moves forward, creating that characteristic curved trajectory.

The key features of any parabola include:

Vertex: The highest or lowest point on the parabola
Axis of symmetry: A vertical line that divides the parabola into two mirror-image halves
Y-intercept: Where the parabola crosses the y-axis
X-intercepts (or zeros): Where the parabola crosses the x-axis

When $a > 0$, the parabola opens upward like a smile 😊, and when $a < 0$, it opens downward like a frown ☹️. The larger the absolute value of $a$, the "narrower" or more "squeezed" the parabola appears.

For the standard form $f(x) = ax^2 + bx + c$, you can find the axis of symmetry using the formula $x = -\frac{b}{2a}$. Once you know this x-value, you can substitute it back into the function to find the y-coordinate of the vertex!

Vertex Form and Transformations

While standard form is great for identifying the y-intercept, vertex form makes it super easy to identify the vertex and understand transformations. The vertex form of a quadratic function is $f(x) = a(x - h)^2 + k$, where $(h, k)$ represents the vertex of the parabola.

Let's break down what each part does:

The value of $a$ still determines whether the parabola opens up or down and how wide it is
The value of $h$ shifts the parabola left or right (opposite of the sign!)
The value of $k$ shifts the parabola up or down

For example, if you have $f(x) = 2(x - 3)^2 + 5$, you can immediately see that:

The parabola opens upward (since $a = 2 > 0$)
The vertex is at $(3, 5)$
It's narrower than the basic parabola $y = x^2$ because $|a| = 2 > 1$

Real-world example: A water fountain shoots water in a parabolic arc. If the fountain's water follows the path $h(t) = -16t^2 + 32t + 6$, where $h$ is height in feet and $t$ is time in seconds, we can convert this to vertex form to find the maximum height. Using our axis of symmetry formula: $t = -\frac{32}{2(-16)} = 1$ second. At $t = 1$: $h(1) = -16(1)^2 + 32(1) + 6 = 22$ feet. So the vertex form would be $h(t) = -16(t - 1)^2 + 22$! 💧

Completing the Square

Sometimes you'll need to convert from standard form to vertex form, and that's where completing the square comes in handy! This algebraic technique is like solving a puzzle - you're trying to create a perfect square trinomial.

Here's the step-by-step process for $f(x) = ax^2 + bx + c$:

Factor out the coefficient of $x^2$ from the first two terms: $f(x) = a(x^2 + \frac{b}{a}x) + c$
Take half of the coefficient of $x$, square it: $(\frac{b}{2a})^2$
Add and subtract this value inside the parentheses
Factor the perfect square trinomial and simplify

Let's try an example: $f(x) = 2x^2 + 12x + 10$

Step 1: $f(x) = 2(x^2 + 6x) + 10$

Step 2: Half of 6 is 3, and $3^2 = 9$

Step 3: $f(x) = 2(x^2 + 6x + 9 - 9) + 10 = 2((x^2 + 6x + 9) - 9) + 10$

Step 4: $f(x) = 2((x + 3)^2 - 9) + 10 = 2(x + 3)^2 - 18 + 10 = 2(x + 3)^2 - 8$

Boom! 💥 Now we can see the vertex is at $(-3, -8)$ and the parabola opens upward.

Projectile Motion Applications

One of the coolest applications of quadratic functions is modeling projectile motion! When you throw, kick, or launch any object through the air (ignoring air resistance), its path follows a parabolic trajectory.

The general equation for projectile motion is: $h(t) = -16t^2 + v_0t + h_0$

Where:

$h(t)$ = height at time $t$ (in feet)
$t$ = time (in seconds)
$v_0$ = initial vertical velocity (in feet per second)
$h_0$ = initial height (in feet)
$-16$ comes from half the acceleration due to gravity ($-32$ ft/s²)

Let's say you're standing on a 50-foot building and throw a ball upward with an initial velocity of 40 ft/s. The equation becomes: $h(t) = -16t^2 + 40t + 50$

To find when the ball hits the ground, set $h(t) = 0$:

$0 = -16t^2 + 40t + 50$

Using the quadratic formula: $t = \frac{-40 ± \sqrt{40^2 - 4(-16)(50)}}{2(-16)} = \frac{-40 ± \sqrt{4800}}{-32}$

This gives us $t ≈ 3.77$ seconds (we ignore the negative solution since time can't be negative in this context).

The maximum height occurs at the vertex. Using $t = -\frac{b}{2a} = -\frac{40}{2(-16)} = 1.25$ seconds, and $h(1.25) = -16(1.25)^2 + 40(1.25) + 50 = 75$ feet! 🏀

Optimization Problems

Quadratic functions are perfect for solving optimization problems - situations where you want to find the maximum or minimum value of something. Since the vertex of a parabola represents either the highest or lowest point, it's exactly what we need! 📊

Here's a classic example: A farmer has 200 feet of fencing and wants to create a rectangular pen with the maximum possible area. If one side of the rectangle is against an existing wall (so he doesn't need fencing for that side), what dimensions should he use?

Let's call the width $x$ feet. Then the length is $(200 - 2x)$ feet (since we need fencing for two widths and one length). The area function becomes:

$A(x) = x(200 - 2x) = 200x - 2x^2 = -2x^2 + 200x$

This is a quadratic function that opens downward (since $a = -2 < 0$), so the vertex gives us the maximum area. Using our formula: $x = -\frac{200}{2(-2)} = 50$ feet.

Therefore, the width should be 50 feet and the length should be $200 - 2(50) = 100$ feet, giving a maximum area of $50 × 100 = 5000$ square feet! 🚜

Another real-world example involves business profits. If a company finds that their daily profit follows the function $P(x) = -2x^2 + 80x - 300$ (where $x$ is the number of items sold), they can find the optimal number of items to sell by finding the vertex: $x = -\frac{80}{2(-2)} = 20$ items, giving a maximum profit of $P(20) = 500$ dollars.

Conclusion

Quadratic functions are mathematical Swiss Army knives! 🔧 We've explored how these powerful functions create parabolic graphs, learned to work with both standard form ($ax^2 + bx + c$) and vertex form ($a(x-h)^2 + k$), mastered the technique of completing the square, and applied our knowledge to real-world scenarios involving projectile motion and optimization problems. Whether you're calculating the perfect angle for a basketball shot, designing the most efficient garden layout, or analyzing business profits, quadratic functions provide the mathematical foundation to solve these problems with confidence and precision.

Study Notes

• Standard Form: $f(x) = ax^2 + bx + c$ where $a ≠ 0$

• Vertex Form: $f(x) = a(x - h)^2 + k$ where $(h, k)$ is the vertex

• Axis of Symmetry: $x = -\frac{b}{2a}$ for standard form, $x = h$ for vertex form

• Parabola Direction: Opens up when $a > 0$, opens down when $a < 0$

• Vertex: Highest point (maximum) when parabola opens down, lowest point (minimum) when opens up

• Projectile Motion Formula: $h(t) = -16t^2 + v_0t + h_0$

• Quadratic Formula: $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

• Completing the Square Steps: Factor out $a$, add/subtract $(\frac{b}{2a})^2$, factor perfect square

• Y-intercept: Found by evaluating $f(0) = c$ in standard form

• Optimization: Maximum/minimum values occur at the vertex of the parabola

• Real-World Applications: Projectile motion, area problems, profit maximization, architectural design