Determining a Quadratic Function When $a = 1$ with Given Vertex or $x$-Intercepts

Introduction

Key Concepts

Understanding Quadratic Functions

A quadratic function is a second-degree polynomial of the form: $$ f(x) = ax^2 + bx + c $$ where $a$, $b$, and $c$ are constants, and $a \neq 0$. The graph of a quadratic function is a parabola, which can open upwards ($a > 0$) or downwards ($a

Vertex Form of a Quadratic Function

The vertex form of a quadratic function reveals its vertex $(h, k)$ and is given by: $$ f(x) = a(x - h)^2 + k $$ When $a = 1$, this simplifies to: $$ f(x) = (x - h)^2 + k $$ The vertex $(h, k)$ represents the highest or lowest point on the parabola, depending on the direction it opens.

Standard Form vs. Vertex Form

While the standard form is useful for identifying the $x$-intercepts and the general shape of the parabola, the vertex form is advantageous for easily locating the vertex. Converting between these forms facilitates various problem-solving techniques.

Determining the Quadratic Function from the Vertex

Given the vertex $(h, k)$ and $a = 1$, the quadratic function can be directly written in vertex form: $$ f(x) = (x - h)^2 + k $$ For example, if the vertex is $(2, -3)$, the function becomes: $$ f(x) = (x - 2)^2 - 3 $$ Expanding this: $$ f(x) = x^2 - 4x + 4 - 3 = x^2 - 4x + 1 $$ Thus, the standard form is $f(x) = x^2 - 4x + 1$.

Finding $x$-Intercepts from the Quadratic Function

The $x$-intercepts (also known as roots or zeros) of a quadratic function are points where $f(x) = 0$. For the standard form $f(x) = x^2 + bx + c$, setting $f(x) = 0$ leads to: $$ x^2 + bx + c = 0 $$ Solving this quadratic equation provides the $x$-intercepts. Methods include factoring, completing the square, or using the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Given $a = 1$, this simplifies to: $$ x = \frac{-b \pm \sqrt{b^2 - 4c}}{2} $$

Example: Determining a Quadratic Function with Given Vertex

**Problem:** Find the quadratic function with vertex at $(3, -2)$ and $a = 1$. **Solution:** Using the vertex form: $$ f(x) = (x - h)^2 + k $$ Substitute $h = 3$ and $k = -2$: $$ f(x) = (x - 3)^2 - 2 $$ Expanding: $$ f(x) = x^2 - 6x + 9 - 2 = x^2 - 6x + 7 $$ Thus, the quadratic function is: $$ f(x) = x^2 - 6x + 7 $$

Example: Determining a Quadratic Function with Given $x$-Intercepts

**Problem:** Find the quadratic function with $x$-intercepts at $x = 1$ and $x = 4$, and $a = 1$. **Solution:** If the $x$-intercepts are $1$ and $4$, the function can be written as: $$ f(x) = (x - 1)(x - 4) $$ Expanding: $$ f(x) = x^2 - 4x - x + 4 = x^2 - 5x + 4 $$ Thus, the quadratic function is: $$ f(x) = x^2 - 5x + 4 $$

Completing the Square

Another method to determine the quadratic function is by completing the square, especially when converting from standard to vertex form or vice versa. This technique involves rewriting the quadratic in a way that reveals the vertex.

Example: Using Completing the Square

**Problem:** Convert the standard form $f(x) = x^2 + 6x + 5$ to vertex form. **Solution:** Start with: $$ f(x) = x^2 + 6x + 5 $$ Complete the square for the $x$ terms: 1. Take half of the coefficient of $x$, which is $3$, and square it to get $9$. 2. Add and subtract $9$ inside the equation: $$ f(x) = (x^2 + 6x + 9) - 9 + 5 = (x + 3)^2 - 4 $$ Thus, the vertex form is: $$ f(x) = (x + 3)^2 - 4 $$ The vertex is $(-3, -4)$.

Discriminant and Nature of Roots

The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by: $$ D = b^2 - 4ac $$ It determines the nature of the roots:

• If $D > 0$, there are two distinct real roots.
• If $D = 0$, there is exactly one real root (a repeated root).
• If $D

Axis of Symmetry

The axis of symmetry of a parabola described by a quadratic function is the vertical line that divides the parabola into two mirror images. Its equation is: $$ x = h $$ where $(h, k)$ is the vertex. This line is pivotal in graphing the quadratic function and understanding its geometric properties.

Graphing Quadratic Functions

To graph a quadratic function, follow these steps:

1. Identify the vertex $(h, k)$.
2. Determine the direction of the parabola (upwards if $a > 0$, downwards if $a
3. Find the axis of symmetry $x = h$.
4. Calculate the $x$-intercepts by setting $f(x) = 0$ and solving for $x$.
5. Determine the $y$-intercept by evaluating $f(0) = c$.
6. Plot these points and sketch the parabola.

Example: Graphing a Quadratic Function

**Problem:** Graph the quadratic function $f(x) = x^2 - 4x + 3$. **Solution:** 1. **Identify Vertex:** Convert to vertex form by completing the square. $$ f(x) = x^2 - 4x + 3 = (x - 2)^2 - 1 $$ Vertex is $(2, -1)$. 2. **Axis of Symmetry:** $x = 2$. 3. **$x$-Intercepts:** Solve $x^2 - 4x + 3 = 0$: $$ (x - 1)(x - 3) = 0 \Rightarrow x = 1, 3 $$ 4. **$y$-Intercept:** $f(0) = 0^2 - 4(0) + 3 = 3$. 5. **Plot Points:** $(2, -1)$, $(1, 0)$, $(3, 0)$, and $(0, 3)$. 6. **Sketch the Parabola:** Draw a symmetric curve passing through these points.

Advanced Concepts

Mathematical Derivations of Quadratic Functions

Understanding the derivation of quadratic functions from different representations enhances comprehension. Starting from vertex form, expanding and rearranging terms aligns it with the standard form, illustrating the interrelation between forms.

Deriving Standard Form from Vertex Form

Given the vertex form: $$ f(x) = (x - h)^2 + k $$ Expanding: $$ f(x) = x^2 - 2hx + h^2 + k $$ Comparing with the standard form $f(x) = x^2 + bx + c$, we identify: $$ b = -2h \\ c = h^2 + k $$ This derivation allows transitioning between forms based on known parameters.

Calculus Connections: Differentiation of Quadratic Functions

In calculus, the differentiation of a quadratic function reveals its rate of change and critical points. For $f(x) = x^2 + bx + c$, the first derivative is: $$ f'(x) = 2x + b $$ Setting $f'(x) = 0$ finds the $x$-coordinate of the vertex: $$ 2x + b = 0 \Rightarrow x = -\frac{b}{2} $$ This aligns with the vertex calculation, emphasizing the link between algebraic methods and calculus.

Applications in Physics: Projectile Motion

Quadratic functions model projectile motion, where the vertical position of an object is a quadratic function of time. For instance: $$ s(t) = -\frac{1}{2}gt^2 + v_0t + s_0 $$ Here, $a = -\frac{1}{2}g$, $b = v_0$, and $c = s_0$. Understanding quadratic functions allows prediction of maximum height, time of flight, and range.

Optimization Problems

Quadratic functions are pivotal in optimization, where maximizing or minimizing a particular quantity is required. For example, determining the dimensions that maximize area given a fixed perimeter involves forming and analyzing a quadratic equation.

Complex Problem-Solving: Multi-Step Quadratic Problems

Advanced problems may integrate quadratic functions with other mathematical concepts. Consider a scenario where a quadratic function intersects another function, requiring simultaneous equations to find points of intersection.

Interdisciplinary Connections: Economics and Revenue Modeling

In economics, quadratic functions model revenue and profit functions. Maximizing profit involves finding the vertex of a quadratic function representing profit against production levels.

Graphical Transformations

Understanding how changes in the function's parameters affect its graph is crucial. Shifts, reflections, and stretching/compressing offer insights into the behavior of quadratic functions under various transformations.

The Quadratic Formula Derivation

Deriving the quadratic formula from the general quadratic equation reinforces the method's validity: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ This formula provides the roots of any quadratic equation and is fundamental in solving quadratic problems.

Discriminant Analysis in Depth

Further exploration of the discriminant $D = b^2 - 4ac$ uncovers deeper insights into the nature of quadratic roots, bridging algebra with complex number theory when $D

Comparison Table

Summary and Key Takeaways