Problem 12 - Algebra and Trigonometry Real Mathematics, Real People

Question

Sketch the graphs of the three functions by hand on the same rectangular coordinate system. Verify your results with a graphing utility.$$\begin{array}{l}f(x)=x^{2} \\\g(x)=\frac{1}{4} x^{2}+2 \\\h(x)=-\frac{1}{4} x^{2}\end{array}$$.

Step-by-Step Solution

Answer

The graphs for \(f(x)=x^2\), \(g(x)=\frac{1}{4}x^2+2\), and \(h(x)=-\frac{1}{4}x^2\) are all parabolas. They all share a vertex at the origin, but \(g(x)=\frac{1}{4}x^2+2\) is shifted 2 units upward. The functions \(f(x)=x^2\) and \(g(x)=\frac{1}{4}x^2+2\) open upwards while \(h(x)=-\frac{1}{4}x^2\) opens downwards. Further, both \(g(x)=\frac{1}{4}x^2+2\) and \(h(x)=-\frac{1}{4}x^2\) are stretched by a factor of \(\frac{1}{4}\) compared to \(f(x)=x^2\).

Step 1: Understand the characteristics of Quadratic Functions

Recognize that the functions are quadratic because they are of the form \(y=ax^2+bx+c\). For \(f(x)=x^2\), the graph will open upwards and the vertex is at the origin because \(a=1\) which is positive and there’s no \(bx\) or \(c\) term to shift the vertex off the origin. For \(g(x) = \frac{1}{4}x^2 + 2\), the graph will also open upwards because \(a=\frac{1}{4}\) which is positive, but the graph is stretched by a factor of \(\frac{1}{4}\) and it's shifted 2 units upwards, making its vertex at (0,2). For \(h(x) = -\frac{1}{4}x^2\), the graph opens downwards because \(a=-\frac{1}{4}\) is negative, indicating an inverted parabola. Its vertex is at the origin.

Step 2: Plotting the Functions

Start with the easier function, \(f(x)=x^2\). It forms a smooth parabolic curve starting from the origin. Continuing with \(g(x) = \frac{1}{4}x^2 + 2\), this graph also has a parabolic shape but it is stretched, meaning it is shallower than \(f(x)\), and its vertex starts from the point (0,2). Finally, for \(h(x) = -\frac{1}{4}x^2\), plot a parabola which opens downwards with the vertex at the origin. It’s also stretched like \(g(x)\), but in the opposite direction.

Step 3: Verify Results with a Graphing Utility

Utilize a graphing utility by entering the equations of the functions and verifying the drawn parabolas correspond to the hand-drawn ones. The graphing software should provide the exact same sketches for each function.

Key Concepts

Graphing Parabolas

Graphing parabolas is a visually intuitive way to understand quadratic functions and how their different forms affect the shape of their graphs. Parabolas are U-shaped curves, and they can either open upwards or downwards based on the coefficient of the squared term. If the coefficient is positive, the parabola opens upwards, and if it’s negative, it opens downwards.

When graphing a parabola by hand, start by identifying the basic form of the quadratic function:

• When given functions like the ones in the exercise: \(f(x) = x^2\), \(g(x) = \frac{1}{4}x^2 + 2\), and \(h(x) = -\frac{1}{4}x^2\), each function provides a certain curvature and direction for the graph.
• Graph \(f(x)=x^2\) by plotting the standard upward opening parabola with its vertex at the origin, (0,0).
• For \(g(x)=\frac{1}{4}x^2+2\), plot the graph by noting that it’s a shallower parabola because the coefficient \(\frac{1}{4}\) stretches it vertically and its vertex starts at (0,2), shifted upwards due to the constant 2.
• Finally, graph \(h(x) = -\frac{1}{4}x^2\) starting from the origin as well but with an inverted curve due to the negative sign.

Using a graphing utility comes as an excellent method to check your hand-drawn graphs and ensure accuracy, as it provides a clear depiction of the parabolic shapes.

Vertex of a Parabola

The vertex of a parabola is a crucial aspect to understand because it is the point where the parabola changes direction. For any given quadratic function \(y = ax^2 + bx + c\), the vertex can usually be found using the formula: \[x = -\frac{b}{2a}\]
This value gives the x-coordinate of the vertex, while you substitute it back into the function to find the y-coordinate.

In our exercise, we can see how different coefficients and constants affect the position of the vertex:

• For \(f(x) = x^2\), the vertex is simply at (0,0) since there are no \(b\) or \(c\) terms to consider.
• In \(g(x) = \frac{1}{4}x^2 + 2\), the vertex is at (0,2) as changing the constant from 0 to 2 shifts it up the y-axis.
• For \(h(x) = -\frac{1}{4}x^2\), it again returns to (0,0) because the parabola is only affected by the negative coefficient leading to a downward open.

Understanding the vertex helps predict where the parabola will sit on the coordinate plane and how transformations alter its placement.

Transformations of Functions

Transformations of functions allow us to manipulate parabolas and move them around the coordinate system without changing their basic shape. These transformations can be vertical shifts, horizontal shifts, vertical stretches or compressions, and reflections.

In our exercise, the functions showcase several transformations:

• The function \(f(x) = x^2\) is the simplest form, presenting a standard parabola centered at the origin.
• For \(g(x) = \frac{1}{4}x^2 + 2\), a vertical shift occurs because the +2 lifts the entire curve upwards. Additionally, having \(\frac{1}{4}\) instead of 1 compresses the parabola vertically, making it appear wider.
• Regarding \(h(x) = -\frac{1}{4}x^2\), the negative leading coefficient results in a reflection across the x-axis, flipping the parabola upside down, while \(\frac{1}{4}\) still stretches it horizontally.

Recognizing these transformations is key to mastering how functions behave and enables you to graph parabolas quickly by anticipating their movement based on changes in the equation.