Problem 13 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Graph the functions by starting with the graph of a familiar function and applying appropriate shifts, flips, and stretches. Label all \(x\) - and \(y\) -intercepts and the coordinates of any vertices and corners. Use exact values, not numerical approximations. (a) \(y=-2(x+1)^{2}+3\) (b) \(y+3=7(x+1)^{2}\)

Step-by-Step Solution

Answer

For function (a), the graph is a reflection of the function \(y=x^2\), stretched by a factor of 2, moved 1 units to the left, and 3 units up. The vertex is at (-1,3), the y-intercept is 1 and the x-intercepts are \(x=-1\pm \sqrt{(3/2)}\). For function (b), the graph is a version the function \(y=x^2\), stretched by factor of 7, moved 1 units to the left, and 3 units down. The vertex is at (-1,-3), the y-intercept is 4 and the x-intercepts are \(x= \pm \sqrt{(3/7)} - 1\).

Step 1: Find the Vertex

The vertex form of a parabola is given by \(y=a(x-h)^{2}+k\), where (h,k) is the vertex of the parabola. For (a) \(y=-2(x+1)^{2}+3\), the vertex is (-1,3) and for (b) \(y=-3+7(x+1)^{2}\), the vertex is (-1,-3).

Step 2: Find the y-intercept

To find the y-intercept, set x = 0. For (a), setting x = 0 gives \(y=-2(0+1)^{2}+3=-2+3=1\), so the y-intercept is 1. For (b), setting x = 0 gives \(y=-3+7(0+1)^{2}=-3+7=4\), so the y-intercept is 4.

Step 3: Find the x-intercepts

To find the x-intercepts, set y = 0. For (a), setting y = 0 gives \(0=-2(x+1)^{2}+3\), solving this equation we get \(x+1=\pm \sqrt{(3/2)}\). Therefore, the x-intercepts are \(x=-1\pm \sqrt{(3/2)}\). For (b), setting y = 0 and solving yields \(0=-3+7(x+1)^{2}\), so \(x= \pm \sqrt{(3/7)} - 1\), are the x-intercepts.

Step 4: Complete the Graph

After plotting the vertices, y-intercepts and x-intercepts on a graph, plot additional points as necessary and construct the parabola through these points.

Key Concepts

Vertex Form

The vertex form of a parabola is a specific way to write the quadratic equation, letting us express a parabola clearly with its vertex at the center. In mathematical terms, it is written as \( y = a(x - h)^2 + k \). Here, \( (h, k) \) represents the vertex of the parabola, which is the point where the parabola changes direction. The value of \( a \) influences how "stretched" or "compressed" the parabola is, along with the direction it opens.
Understanding the vertex form makes it easier to graph parabolas, as you can directly identify the vertex from the equation without converting it from the standard form. For example, for the equation \( y = -2(x + 1)^2 + 3 \), the vertex is \( (-1, 3) \), because \( x + 1 \) equals zero when \( x = -1 \), and thus \( y = 3 \). This quick identification is helpful in graphing the parabola accurately.

X-intercepts

To find where a parabola crosses the x-axis, you need to find the x-intercepts. These are the points where the output of the quadratic function is zero, meaning \( y = 0 \). For solving this, you must set the equation equal to zero and solve for \( x \).
For instance, consider the equation \( y = -2(x + 1)^2 + 3 \). Setting \( y = 0 \) gives \( -2(x + 1)^2 + 3 = 0 \). Solving for \( x \) involves rearranging to find \( x + 1 = \pm \sqrt{\frac{3}{2}} \). Hence, the x-intercepts are \( x = -1 \pm \sqrt{\frac{3}{2}} \). This solution shows that the parabola does not intersect at single points, but rather at values that can be calculated to see the actual roots.

Y-intercepts

Y-intercepts are the intersection points of a graph with the y-axis, where the value of \( x \) is zero. Calculating these helps to quickly place one of the essential points on the graph for sketching the parabola.
To find this intercept for any quadratic in vertex form, substitute \( x = 0 \) into the equation and solve for \( y \). For example, for the equation \( y = -2(x + 1)^2 + 3 \), set \( x = 0 \):

• \( y = -2(0 + 1)^2 + 3 \)
• \( y = -2 + 3 \)
• Thus, \( y = 1 \)

Therefore, the y-intercept is 1. Knowing this allows you to start plotting from the vertex and proceed with creating the parabolic curve.

Parabola Transformations

Parabolas can undergo various transformations that shift or deform the graph while retaining its parabolic shape. Understanding these transformations can aid in visualizing and graphing the functions accurately.
Here are the main types of transformations:

• Vertical Shifts: Adjust the k value in the vertex form \((y = a(x-h)^2 + k)\) to move the parabola up or down along the y-axis. Positive k shifts it upwards, while negative k shifts it downwards.
• Horizontal Shifts: Modify the h value. If \( h \) is positive, the graph moves to the left; if negative, it shifts to the right.
• Vertical Stretch/Compression and Reflection: Change in the \( a \) value affects the parabola's "width" and direction. An \( |a| > 1 \) stretches the parabola, while \( 0 < |a| < 1 \) compresses it. Negative \( a \) indicates the parabola opens downwards, reflecting it across the x-axis.

By applying these transformation principles, you can easily transform and accurately sketch a variety of parabolas.