Problem 13 - Algebra 1: Concepts and Skills

Question

Sketch the graph of \(y=(x+2)(x-2) .\) Label the vertex and the \(x\) -intercepts.

Step-by-Step Solution

Answer

The \(x\)-intercepts are \((-2,0)\) and \((2,0)\) and the vertex is \((0,-4)\). The graph is a parabola opening upward.

Step 1: Find the x-intercepts

To find the x-intercepts, you need to set \(y=0\) and solve for \(x\). This means solving the equation \((x+2)(x-2)=0\). The solutions to this equation are \(x=-2\) and \(x=2\), so these are the \(x\)-intercepts.

Step 2: Find the vertex

Since the parabola opens upward, the vertex of the parabola is the lowest point on the graph. The x-value of the vertex is the average of the \(x\)-intercepts, which is \(\frac{-2+2}{2}=0\). Plugging \(x=0\) into the equation for the graph gives \(y=(0+2)(0-2)=(2)(-2)=-4\). So, the vertex is at the point \((0, -4)\).

Step 3: Sketch the graph

Start by plotting the \(x\)-intercepts \((-2,0)\) and \((2,0)\) and the vertex \((0,-4)\). Draw a smooth, U-shaped curve through these points. The curve should open upwards because the coefficient of \(x^2\) is positive. Label the vertex and the \(x\)-intercepts.

Key Concepts

X-Intercepts

Understanding x-intercepts is key to graphing quadratic equations. In a quadratic graph, x-intercepts are where the parabola crosses the x-axis. Mathematically, this is where the value of y is zero. To find these intercepts, you set the equation y = 0 and solve for x. This usually involves factoring the quadratic equation or using the quadratic formula. In the example y=(x+2)(x-2), the solutions to the equation are x=-2 and x=2, after setting y=0. Those points, (-2, 0) and (2, 0), are the x-intercepts of the parabola and crucial for sketching its overall shape.

Vertex of a Parabola

The vertex of a parabola is the peak or the lowest point depending on whether the parabola opens downwards or upwards, respectively. For the latter, the vertex represents the minimum point. To find the vertex, you need the x-coordinates of the x-intercepts. They help in determining the line of symmetry of the parabola, which passes through the vertex. The vertex's x-value is the midpoint of the x-intercepts. In the exercise, by averaging the x-intercepts x=-2 and x=2, we get x=0. Substituting this value into the original equation gives the y-coordinate as y=-4, which makes the vertex (0, -4). This point is significant as it helps to accurately plot the parabola and reflects its axis of symmetry.

Quadratic Functions

Quadratic functions are fundamental in algebra and are expressed in the standard form y = ax^2 + bx + c. These equations create parabolic graphs that can open up or down, based on the sign of coefficient a. The graph's shape is symmetrical, and has a single vertex. The importance of understanding the structure of quadratic functions lies in their wide range of applications, from physics to finance. Recognizing how changes to the values of a, b, and c will affect the graph's shape enables us to analyze and predict behavior in various scenarios. The given equation y=(x+2)(x-2) simplifies to y=x^2-4, which is a parabola that opens upward due to the positive x^2 term. Through this function, you can see how the vertex and x-intercepts play a role in graphing the entire parabola.