Problem 57 - Algebra and Trigonometry Real Mathematics, Real People

Question

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

Step-by-Step Solution

Answer

The upward-opening quadratic function is \(f(x) = (x+1.75)^2 - 1.75^2\) and the downward-opening function is \(g(x) = -(x+1.75)^2 - 1.75^2\), where \(x\) is in the domain of all real numbers.

Step 1: Find the vertex

As the x-intercepts are symmetrically situated about the vertex, the horizontal coordinate \(h\) of the vertex will be midway between the x-intercepts \(-3\) and \(-\frac{1}{2}\). To find \(h\), average these two x-values: \( h = \frac{-3 + (-0.5)}{2} = -1.75\).

Step 2: Set up the two quadratic functions

The vertical coordinate \(k\) of the vertex is not directly given or needed, as the x-intercepts will be the same for any value of \(k\). So, arbitrarily set \(k = 0\). Also, the scale factor \(a\) needs to be positive for the parabola to open upward and negative for the parabola to open downward. Arbitrarily set \(a = 1\) for the upward function and \(a = -1\) for the downward function. So, the upward-opening function is \(f(x) = (x+1.75)^2\), and the downward-opening function is \(g(x) = -(x+1.75)^2\).

Step 3: Check the x-intercepts

To validate, check if they generate the given x-intercepts when set equal to zero. Solving \(f(x) = 0\) yields \(x = -1.75 \pm \sqrt{0} = -1.75\). So, the only x-intercept of \(f(x)\) is at \(-1.75\), which is incorrect. Similarly, the only x-intercept of \(g(x)\) is also at \(-1.75\), which is also incorrect.

Step 4: Adjust the vertex

As Step 3 failed, instead of arbitrarily setting the vertex to \((-1.75, 0)\), it seems that the best choice for \(h\) is the midpoint between the two roots, and \(k = 0\). So, \(h = \frac{-3 - (-0.5)}{2} = -1.75\), \(k = 0\). Now the quadratic functions become \(f(x) = (x+1.75)^2 - (-1.75)^2\), \(g(x) = -(x+1.75)^2 - (-1.75)^2\).

Step 5: Check the x-intercepts again

Now check for x-intercepts by setting the functions to zero: For \(f(x)\), this gives \(x = -1.75 \pm \sqrt{(-1.75)^2} = -3, -0.5\), and for \(g(x)\) the same. Now both functions have the correct x-intercepts of \(-3\) and \(-\frac{1}{2}\).

Key Concepts

x-intercepts

When understanding quadratic functions, identifying the x-intercepts is crucial. These are the points where the graph of the function crosses the x-axis. In mathematical terms, x-intercepts are the solutions to the equation when it is set to zero, i.e., where the y-value is zero: \( f(x) = 0 \). For the quadratic formula \( ax^2 + bx + c = 0 \), the x-intercepts can be found using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This requires calculating the discriminant, \( b^2 - 4ac \). If positive, there are two x-intercepts; if zero, one; and if negative, none.
Understanding the placement of x-intercepts helps in sketching the parabola and solving for other properties. In our specific exercise, the provided x-intercepts were

• \(x = -3\)
• \(x = -\frac{1}{2}\)

The task was to construct quadratic functions that include these intercepts, ensuring accurate representation on the graph.

vertex of a parabola

The vertex of a parabola is a key feature representing the highest or lowest point, depending on the parabola's direction. For the standard quadratic function \( y = a(x-h)^2 + k \), the point \((h, k)\) is the vertex.

• The horizontal coordinate \(h\) is exactly midway between the x-intercepts.
• The vertical coordinate \(k\) adjusts the parabola's height but is not essential for determining the x-intercepts.

In the exercise, we deduced \(h\) by averaging the x-intercepts \[ h = \frac{-3 + -\frac{1}{2}}{2} = -1.75 \]. Choosing \(k = 0\), we simplified our calculations while ensuring the quadratic function's vertex lays directly between the intercepts, aligning symmetrically.

parabola direction

The direction of the parabola is determined by the coefficient \(a\) in the quadratic equation \( y = a(x-h)^2 + k \). This factor influences the parabola's opening:

• If \(a > 0\), the parabola opens upwards, like a smile.
• If \(a < 0\), it opens downwards, resembling a frown.

This characteristic was particularly useful in constructing two separate functions in the exercise. We chose an

• upwards opening with \(a = 1\) for one quadratic function
• downwards opening with \(a = -1\) for the other

By manipulating \(a\), we controlled the parabola's direction while maintaining the same x-intercepts, thereby creating distinct variations of quadratic functions using the same basic parameters.

finding quadratic equations

Finding a quadratic equation involves piecing together several components: x-intercepts, the vertex, and the direction of the parabola. Here’s a simplified method:- **Determine the x-intercepts:** These form the basis of the quadratic function where solutions exist.- **Find the vertex:** Calculate the mean of the x-intercepts for the horizontal component and choose an appropriate vertical position if needed.- **Decide on the direction:** Use \(a\) to define the direction (upwards or downwards).In our example, we built two functions:

• \(f(x) = (x + 3)(2x + 1)\) for upwards and
• \(g(x) = -(x + 3)(2x + 1)\) for downwards

Each function shares the same x-intercepts. This approach shows how the relationship between these components is flexible and allows the creation of various quadratic equations based on specified conditions.