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Chapter 2: Problem 70

Find two quadratic functions, one that opens upward and one that opens downward, whose graphs have the given \(x\) -intercepts. (There are many correct answers.) \(\left(-\frac{5}{2}, 0\right),(2,0)\)

Short Answer

Expert verified
The solutions indicating two quadratic equations given the x-intercepts are: For upward facing parabola: \(f(x) = (x + \frac{1}{4})^2\), and for downward facing parabola: \(f(x) = -(x + \frac{1}{4})^2\)

Step by step solution

01

- Calculate vertex x-coordinate

First, we will find the x-coordinate of the vertex of the quadratic equation using the given x-intercepts (or roots). It is the average of the roots, i.e. \(h = \frac{-5/2 + 2}{2} = -\frac{1}{4}\)
02

- Formulate quadratic equations

Now we will generate two quadratic equations using the vertex form \(f(x) = a(x - h)^2 + k\), where \(a\), \(h\) and \(k\) are constants. The quadratic function that opens upward will have \(a > 0\) and the one that opens downward will have \(a < 0\). We also know that because the parabolas cross the x-axis at \(-5/2\) and \(2\), the value of \(k\) (the y-coordinate of the vertex) will be 0. Therefore we can write the two functions as:\n\n1. Upward facing parabola: \(f(x) = a(x + \frac{1}{4})^2\)\n2. Downward facing parabola: \(f(x) = -a(x + \frac{1}{4})^2\)\n\nHere, \(a\) can be any positive number for the upward facing parabola and any negative number for the downward facing parabola.
03

- Finalize the functions

Now we simply need to choose an \(a\) value for each equation. As there are many correct answers, it's reasonable to choose \(a = 1\) for the upward facing parabola and \(a = -1\) for the downward facing parabola, making the solutions easeier to interpret.\n\nThe two quadratic functions are:\n\n1. Upward facing parabola: \(f(x) = (x + \frac{1}{4})^2\)\n2. Downward facing parabola: \(f(x) = -(x + \frac{1}{4})^2\)

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