Problem 43 - 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.) $$(0,0),(10,0)$$

Step-by-Step Solution

Answer

The two quadratic functions are \(f(x) = (x-5)^2\) for upward opening and \(g(x) = -(x-5)^2\) for downward opening.

Step 1: Identify the vertex

The vertex is halfway between the x-intercepts, so in this case, \(h = 5\) and \(k = 0\). Thus, the quadratic function takes the form \(f(x) = a(x-5)^2 + 0\).

Step 2: Construct the functions for upward and downward openings

To make the function open upwards, choose \(a = 1\). So, the function is \(f(x) = (x-5)^2\). To make the function open downwards, choose \(a = -1\). So, the function is \(g(x) = -(x-5)^2\).

Key Concepts

X-Intercepts

Understanding x-intercepts is crucial when dealing with quadratic functions. The x-intercepts are the points where the graph of a quadratic function crosses the x-axis. Algebraically, these are the values of x for which the quadratic equation equals zero. In the context of our example, the x-intercepts are given as \( (0,0) \), and \( (10,0) \).

To find a quadratic equation with these intercepts, we can use the fact that the quadratic function \( f(x) = a(x - x_1)(x - x_2) \) will have x-intercepts at \( x_1 \) and \( x_2 \) if \( a \) is not zero. Here, our x-intercepts \( x_1 = 0 \) and \( x_2 = 10 \) can be plugged into the standard form to begin constructing our functions.

Vertex of a Parabola

The vertex of a parabola is a significant point that shows the maximum or minimum value of the function, depending on its orientation. It is the point where the parabola changes direction. For our example, the vertex is found by taking the average of the x-intercepts, which in this case is \( h = 5 \) and \( k = 0 \), forming the vertex \( (5,0) \).

The standard form of a quadratic function is \( f(x) = a(x - h)^2 + k \) where \( (h,k) \) is the vertex of the parabola. Therefore, we construct the quadratic functions using the identified vertex, ensuring that we have accommodated the correct positioning on the graph.

Upward Opening Parabola

An upward opening parabola takes the shape of a 'U' and it indicates that the leading coefficient \( a \) of the quadratic function is positive. For a function to open upward, any positive value can be assigned to \( a \), and in the simplest case we often choose \( a = 1 \).

In our exercise example, when \( a = 1 \), the quadratic function \( f(x) = (x-5)^2 \) is an upward opening parabola with the previously discussed vertex and x-intercepts. It will have a minimum value at the vertex point since the parabola is opening upwards.

Downward Opening Parabola

Conversely, a downward opening parabola resembles an upside-down 'U', created when the leading coefficient \( a \) is negative. Setting \( a = -1 \) will make the parabola open downwards. This reflects a quadratic function with a maximum value at its vertex.

Referring back to the exercise, by choosing \( a = -1 \) we get the function \( g(x) = -(x-5)^2 \) which opens downwards. This means the graph arcs downward from the vertex, reaching its maximum height at the vertex before descending towards the intercepts on the x-axis.