Problem 45 - 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.) $$(-3,0),\left(-\frac{1}{2}, 0\right)$$

Step-by-Step Solution

Answer

The two quadratic functions that satisfy the given conditions are \(f_1(x) = 1(x-h)^2\) and \(f_2(x) = -1(x-h)^2\), where \(h\) is calculation from Step 1.

Step 1: Determine the vertex

The x-coordinate of the vertex (h) is the average of the x-coordinates of the given intercepts. So, calculate \(h = \frac{(-3) + (-\frac{1}{2})}{2}\).

Step 2: Create an upward-opening quadratic function

Arbitrarily choose a positive value for a. Since we want the graph to open upwards, a should be positive. We can take \(a = 1\), for simplicity. The quadratic function will be \(f_1(x) = 1(x-h)^2\).

Step 3: Create a downward-opening quadratic function

Arbitrarily choose a negative value for a. Since we want the graph to open downwards, a should be negative. We can take \(a = -1\), for simplicity. The quadratic function will be \(f_2(x) = -1(x-h)^2\).

Key Concepts

x-intercepts

The x-intercepts of a quadratic function are the points where the graph crosses the x-axis. In simpler terms, these are the points where the value of the function is zero. For our given exercise, these intercepts are

• (-3, 0)
• \((-\frac{1}{2}, 0)\)

These x-intercepts suggest that the roots of the quadratic equation derived from this function are -3 and \(-\frac{1}{2}\).
This is important because, for any quadratic function of the form \[f(x) = a(x-r_1)(x-r_2)\]the values \(r_1\) and \(r_2\) represent the x-intercepts of the function. Understanding how to find the x-intercepts is fundamental to graphing and solving quadratic equations.

vertex

The vertex of a quadratic function, represented by \((h, k)\), is the point where the parabola changes direction. It's a turning point on the graph.
To find the x-coordinate of the vertex for our quadratic function with the intercepts -3 and \(-\frac{1}{2}\), we need to calculate the average of these x-coordinates. This can be calculated as follows:\[h = \frac{(-3) + (-\frac{1}{2})}{2}\]After performing the calculation, you find the x-coordinate of the vertex. The y-coordinate \(k\) can be found by inserting \(h\) back into the quadratic equation or observed in cases where the parabola sits on the x-axis at the intercepts.
The vertex serves as a crucial point for defining the shape and position of the parabola on a graph. It helps to understand whether the parabola will be vertically symmetrically placed and gives insights into the maximum and minimum values of quadratic functions.

upward and downward opening

The direction in which a parabola opens is determined by the sign of the coefficient \(a\) in the quadratic function \[f(x) = a(x-h)^2 + k\].
If \(a\) is positive, the parabola opens upwards, resembling a U-shape. Conversely, if \(a\) is negative, the parabola opens downwards, appearing like an upside-down U. Understanding this concept is essential to predict the nature and shape of the quadratic function's graph.

• Upward-Opening:
  When \(a=1\), the function takes the form \[f_1(x) = 1(x-h)^2\], signifying an upward-opening parabola.
• Downward-Opening:
  When \(a=-1\), the function becomes \[f_2(x) = -1(x-h)^2\], depicting a downward-opening parabola.

This understanding helps evaluate whether the function will have a maximum or a minimum point and is crucial for solving real-world problems where such calculations matter.

average of coordinates

In the context of quadratic functions, particularly when determining the vertex, the average of the coordinates plays a key role. Specifically, the x-coordinate of the vertex is found by taking the average of the x-values of the intercepts. This is an essential property because it makes finding the vertex more straightforward.

• Determine the midpoint or average by adding the two intercepts and dividing by 2.
• For our given intercepts: \(-3\) and \(-\frac{1}{2}\), calculate the average:\[h = \frac{(-3) + (-\frac{1}{2})}{2}\]This will provide the x-coordinate \(h\) of the vertex.

This technique simplifies the graphing of a parabola and provides a clear understanding of where the "turning point" of the function will lie on the x-axis. Knowing this concept helps manage and predict the behavior of quadratic equations in more practical and applicable scenarios.