Problem 23 - Elementary Algebra

Question

Graph. Find the vertex and the y-intercept. In addition, find the \(x\) - intercepts if they exist. $$ y=x 2-2 x-8 $$

Step-by-Step Solution

Answer

Vertex: (1, -9); y-intercept: (0, -8); x-intercepts: (4, 0) and (-2, 0).

Step 1: Identify the type of equation

The given equation is a quadratic equation in the form of \(y = ax^2 + bx + c\), where \(a = 1\), \(b = -2\), and \(c = -8\). Quadratic equations form parabolas when graphed.

Step 2: Find the vertex

The vertex of a parabola in the form \(y = ax^2 + bx + c\) can be found using the formula \(x = -\frac{b}{2a}\). First, calculate \(x = -\frac{-2}{2(1)} = 1\). Then substitute \(x = 1\) back into the equation to find \(y\): \(y = (1)^2 - 2(1) - 8 = 1 - 2 - 8 = -9\). So, the vertex is \((1, -9)\).

Step 3: Find the y-intercept

The y-intercept occurs where \(x = 0\). Substitute \(x = 0\) into the equation: \(y = (0)^2 - 2(0) - 8 = -8\). So, the y-intercept is \((0, -8)\).

Step 4: Find the x-intercepts

To find the x-intercepts, set \(y = 0\) and solve the equation \(x^2 - 2x - 8 = 0\). Factor the equation: \((x - 4)(x + 2) = 0\). Set each factor equal to zero: \(x - 4 = 0\) or \(x + 2 = 0\), giving solutions \(x = 4\) and \(x = -2\). Therefore, the x-intercepts are \((4, 0)\) and \((-2, 0)\).

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a crucial point representing its peak or trough on the graph, depending on the parabola's orientation. For a standard quadratic equation in the form \( y = ax^2 + bx + c \), the vertex can be found using the formula:

• \( x = -\frac{b}{2a} \)

This formula helps locate the x-coordinate of the vertex. In our example, with the equation \( y = x^2 - 2x - 8 \), we can calculate \( x = -\frac{-2}{2 \times 1} = 1 \). To find the corresponding y-coordinate, substitute this x value back into the equation:

• \( y = 1^2 - 2 \times 1 - 8 = -9 \)

Thus, the vertex is \( (1, -9) \). This point is the minimum point of the parabola because the parabola opens upwards (as the coefficient of \( x^2 \) is positive).

Y-Intercept

The y-intercept of a quadratic equation is found where the graph crosses the y-axis. This occurs when \( x = 0 \), allowing us to easily determine the y-coordinate by substituting \( x = 0 \) into the equation. For our given equation:

• \( y = 0^2 - 2 \times 0 - 8 = -8 \)

Hence, the y-intercept of the parabola is \( (0, -8) \). This point gives us a starting value when plotting the graph and provides important insight into the equation's behavior as \( x = 0 \). It is a signature feature of the parabola as it really shows how the graph relates to the y-axis.

X-Intercepts

The x-intercepts are the points where the graph of the equation crosses the x-axis. These occur where the y-value is zero. For a quadratic equation like \( y = x^2 - 2x - 8 \), we set \( y = 0 \) and solve:

• \( x^2 - 2x - 8 = 0 \)

This equation can be factored into \( (x - 4)(x + 2) = 0 \). From these factors, we can find the x-values by setting each factor equal to zero:

• \( x - 4 = 0 \) leads to \( x = 4 \)
• \( x + 2 = 0 \) leads to \( x = -2 \)

Thus, the x-intercepts are located at \( (4, 0) \) and \( (-2, 0) \). These points are vital as they show where the graph touches or crosses the x-axis, showing the roots of the equation.

Parabolas

A parabola is a U-shaped curve that can open either upwards or downwards, determined by the sign of the coefficient of \( x^2 \) in a quadratic equation. If the coefficient is positive, like in \( y = x^2 - 2x - 8 \), the parabola opens upwards, indicating a minimum vertex. Conversely, if it were negative, the parabola would open downwards, indicating a maximum vertex.
Parabolas have key features including vertices, axes of symmetry, and intercepts:

• The vertex is either the lowest or highest point.
• The y-intercept represents where the parabola crosses the y-axis.
• The x-intercepts are where the graph touches or intersects with the x-axis.

Understanding parabolas through these features not only aids in graphing but also in solving real-world problems involving quadratic equations.

Factoring Quadratic Equations

Factoring quadratic equations is an essential skill, especially when finding x-intercepts. This process involves rewriting the quadratic equation in the form \( ax^2 + bx + c = (px + q)(rx + s) \), making it easier to determine solutions.

To factor the equation \( x^2 - 2x - 8 = 0 \), we look for two numbers that multiply to \( -8 \) (the constant term) and add to \( -2 \) (the linear coefficient). In this case, those numbers are \( -4 \) and \( 2 \), giving us the factorization:

• \( (x - 4)(x + 2) = 0 \)

Once factored, setting each binomial equal to zero reveals the solutions for \( x \), namely \( x = 4 \) and \( x = -2 \). Factoring not only gives solutions for x-intercepts but simplifies solving quadratics by hand, enhancing computational efficiency.