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Chapter 2: Problem 69

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

Short Answer

Expert verified
The quadratic function that opens upwards is \(f(x) = x^2 + \frac{7}{2}x +\frac{3}{2}\) and the one that opens downwards is \(f(x) = -x^2 - \frac{7}{2}x -\frac{3}{2}\)

Step by step solution

01

Identify the x-intercepts

The given x-intercepts are \(-3\) and \(-\frac{1}{2}\), which are the roots of the quadratic equation and can be represented as \(x+3=0\) and \(x+\frac{1}{2}=0\), respectively.
02

Formulate Quadratic Functions

The standard form of a quadratic function with roots \(x_1\) and \(x_2\) is \(f(x) = a(x - x_1)(x - x_2)\). To find a quadratic function that opens upwards, choose \(a=1\), and to find one that opens downwards, choose \(a=-1\).
03

Apply the x-intercepts

By substituting \(x_1= -3\) and \(x_2=-\frac{1}{2}\) for both functions, we get \(f(x) = (x + 3)(x + \frac{1}{2})\) for the function that opens upwards, and \(f(x) = - (x + 3)(x + \frac{1}{2})\) for the function that opens downwards.
04

Simplify the Quadratic Functions

Finally simplify both equations to get their final forms. For the function that opens upwards it becomes \(f(x) = x^2 + \frac{7}{2}x +\frac{3}{2}\), and for the one that opens downwards \(f(x) = -x^2 - \frac{7}{2}x -\frac{3}{2}\).

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