Problem 46 - 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.) $$\left(-\frac{5}{2}, 0\right),(2,0)$$

Step-by-Step Solution

Answer

The two quadratic functions whose graphs have the x-intercepts of \(-5/2\) and \(2\) with one function opening up and the other opening down can be \(f_1(x) = (x + \frac{5}{2})(x - 2)\) and \(f_2(x) = -(x + \frac{5}{2})(x - 2)\), respectively.

Step 1: Form the basic equations

Substitute the given x-intercepts into the equation to obtain the basic forms of the quadratic functions. So the basic forms will be \(f_1(x) = a(x - (-5/2))(x - 2)\) for the function that opens up and \(f_2(x) = -a(x - (-5/2))(x - 2)\) for the function that opens down, where \(a > 0\).

Step 2: Simplify the Equations

Simplify the equations \(f_1(x)\) and \(f_2(x)\). This yields the functions \(f_1(x) = a(x + \frac{5}{2})(x - 2)\) and \(f_2(x) = -a(x + \frac{5}{2})(x - 2)\).

Step 3: Specify Value for a

Choose a value for \(a\) in both equations. Remembering that \(a\) must be \(>0\), let's choose \(a = 1\) as an example. This will yield the two specific quadratic functions \(f_1(x) = (x + \frac{5}{2})(x - 2)\) and \(f_2(x) = -(x + \frac{5}{2})(x - 2)\).

Key Concepts

x-intercepts of a quadratic

Understanding the x-intercepts of a quadratic function can be quite illuminating, as these are the points where the graph of the function crosses the x-axis, meaning the output, or y-value, is zero at these points. To find them, one simply sets the quadratic function equal to zero and solves for x. For the quadratic functions provided, the x-intercepts are given as \(\left(-\frac{5}{2}, 0\right)\) and \(\left(2, 0\right)\), and these intercepts form the foundation from which we build the rest of the function.

For a general quadratic function \(f(x) = ax^2 + bx + c\), the x-intercepts can be found analytically using the quadratic formula. However, when specific x-intercepts are known, as in our exercise, we can represent the function as \(f(x) = a(x - x_1)(x - x_2)\), where \(x_1\) and \(x_2\) are the intercepts. This form is particularly helpful because it makes it clear that if you plug in \(x_1\) or \(x_2\) for x, the output is zero, directly showing their significance as intercepts.

graph of a quadratic equation

The graph of a quadratic equation is a symmetrical curve called a parabola. A unique feature of this graph is that it can either open upwards or downwards, which is determined by the coefficient of the \(x^2\) term in the quadratic equation. If we consider the quadratic equation in its standard form \(f(x) = ax^2 + bx + c\), the sign of the coefficient a decides the direction in which the parabola opens.

In our exercise, we have two quadratic functions with the same x-intercepts, yet they differ in the way their graphs open. The graph of \(f_1(x)\) will open upwards because the leading coefficient is positive, while the graph of \(f_2(x)\) will open downwards due to its negative leading coefficient. The vertex of a parabola, the highest or lowest point depending on the opening direction, is also a crucial concept. For a quadratic function in standard form, the vertex's x-coordinate is given by \(x = -\frac{b}{2a}\), and you can find the y-coordinate by substituting this back into the function.

quadratic equations opening up or down

Whether a quadratic equation's graph opens up or down is central to understanding its behavior. The key is the value of the coefficient a in the standard form of the quadratic equation \(f(x) = ax^2 + bx + c\). When a is positive, the parabola opens upward, meaning it has a minimum point at its vertex; conversely, a negative a results in the parabola opening downward, indicating a maximum point at the vertex.

This characteristic affects not only the graph's shape, but also the range of the quadratic function. For an upward-opening parabola, the range includes all y-values greater than or equal to the y-coordinate of the vertex. For a downward-opening parabola, the range comprises all y-values less than or equal to the vertex's y-coordinate. In the solutions presented, we chose a positive value for a to create \(f_1(x)\) that opens upward, and a negative value to create \(f_2(x)\) for the downward-opening case, highlighting the impact of the a coefficient's sign on the graph's direction.