Problem 44 - College Algebra and Calculus: An Applied Approach

Question

Find two quadratic functions whose graphs have the given \(x\) -intercepts. Find one function whose graph opens upward and another whose graph opens downward. (There are many correct answers.) $$(4,0),(8,0)$$

Step-by-Step Solution

Answer

A function that opens upward is given by \(y = (x - 4)(x - 8)\) and a function that opens downward is given by \(y = -(x - 4)(x - 8)\). Note that any positive or negative number may replace 1 to generate a multitude of correct answers to the problem.

Step 1: Compute the vertex of the parabola

The vertex of the quadratic function will be at the midpoint between the x-intercepts. The midpoint is calculated by taking the average of the x-intercepts, which in this case is \((4 + 8) / 2 = 6\). Therefore, the vertex \(h\) is at (6, 0).

Step 2: Substitute the vertex into the quadratic function

With the vertex identified as (6,0), the form of the quadratic function becomes \(y=a(x-6)^2\). Given that we want the function to pass through intercepts (4,0) and (8,0), we know that for these points \(y = 0\). To solve for \(a\), any one of those points can be used, as we know both \(x\) and \(y\). Let's take (4,0). Substitute these values into the equation, which gives \(0 = a(4-6)^2\). Solving this equation for \(a\) will provide the value needed.

Step 3: Solve for \(a\)

Solving the equation \(0 = a(4-6)^2\) for \(a\) yields \(a= 0\). However, this would imply a straight line, not a parabola, which contradicts the premise of the problem. The contradiction reveals that the correct form of the quadratic function must be \(y = a(x - h)(x - k)\). So, the form of the quadratic function now becomes \(y = a(x - 4)(x - 8)\). Set \(y = 0\) to find \(a\).

Step 4: Solution for the quadratic functions

With the correct quadratic form \(y = a(x - 4)(x - 8)\), we have infinite correct answers. The problem asked for a parabola that opens upwards, which means \(a > 0\), and for a parabola that opens downwards, which means \(a < 0\). Therefore, any positive value for \(a\) will define a parabola that opens upwards, and any negative value for \(a\) will define a parabola that opens downwards. Taking \(a = 1\) and \(a = -1\) for simplicity, the quadratic functions are \(y = (x - 4)(x - 8)\) and \(y = -(x - 4)(x - 8)\) respectively.

Key Concepts

X-Intercepts

To understand x-intercepts in the context of quadratic functions, think of them as the points where the graph of the parabola crosses the x-axis. In the given exercise, the x-intercepts are provided as (4, 0) and (8, 0).

For any parabola, the x-intercepts can be found where the value of the function is zero, which is why the y-coordinate of these points is always zero. These intercepts hold essential information about the parabola's shape and orientation. When we have two distinct x-intercepts, we can conclude that we have a 'real' and 'non-degenerate' parabola, which means it's not just a straight line.

Finding the quadratic functions based on these intercepts involves putting the equation into factored form:

• Standard form: y = ax^2 + bx + c
• Factored form: y = a(x - x1)(x - x2), where x1 and x2 are the x-intercepts

In our case, knowing these two points, the factored form becomes y = a(x - 4)(x - 8), with 'a' determining the direction the parabola opens.

Vertex of a Parabola

The vertex of a parabola is the peak or the lowest point of the curve, depending on whether the graph opens upward or downward, respectively.

In the exercise, the vertex is found by calculating the midpoint between the x-intercepts (4, 0) and (8, 0), which gives us the vertex (6, 0). This position is crucial because it gives us the maximum or minimum value of the quadratic function, depending on the orientation of the parabola.

• For an upward-opening parabola, the vertex is at the minimum point.
• For a downward-opening parabola, the vertex is at the maximum point.

The vertex form of a quadratic function is expressed as y = a(x - h)^2 + k, where (h, k) are the coordinates of the vertex. In our exercise, where the vertex is (6, 0), the vertex form would be y = a(x - 6)^2, with 'a' determining the 'width' and 'direction' of the parabola opening.

Solving Quadratic Equations

The process of solving quadratic equations is essential to find the x-intercepts of its graph. A quadratic equation can typically be written as ax^2 + bx + c = 0.

To solve for the x-intercepts, also known as 'roots', you can employ several methods, such as factoring, completing the square, or using the quadratic formula.

In this exercise, after determining the vertex and understanding that the two x-intercepts must satisfy the quadratic equation for y = 0, we set up y = a(x - 4)(x - 8) to find 'a'. Notably, 'a' affects the orientation of the parabola. The process of finding 'a' is essentially solving for the coefficient that brings the function down to the x-axis at the appointed intercepts.

Choosing a positive 'a' gives an upward opening parabola, while a negative 'a' yields a parabola that opens downward. It's a reminder that the solutions to quadratic equations are not just about finding numerical values but also understanding their graphical meanings in the context of the parabola's shape and direction.