Problem 6 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For Problems 1 through 8, graph the function. Label the \(x\) - and \(y\) -intercepts and the coordinates of the vertex. $$ f(x)=-3(x+2)^{2}+9 $$

Step-by-Step Solution

Answer

The graph of the function \( f(x)=-3(x+2)^{2}+9 \) is a downward-opening parabola with vertex at (-2, 9) and y-intercept at (0, -3). It does not have x-intercepts.

Step 1: Identify the Vertex of the Function

The function \( f(x)=-3(x+2)^{2}+9 \) is already in vertex form. The vertex is given by the coordinates \((h, k)\). Therefore, comparing this to the standard vertex form, we find the vertex is given by the coordinates \((-2, 9)\).

Step 2: Find the y-intercept of the Function

To find the y-intercept, we set \(x = 0\) in the equation and find the value of \(y\). Plugging \(x = 0\) into \( f(x)=-3(x+2)^{2}+9 \) results in \( f(0)=-3(0+2)^{2}+9 = -3 \). So, the y-intercept is at the point \((0, -3)\).

Step 3: Find the x-intercepts of the Function

To find the x-intercept, we need to set \(f(x)=0\). Solving the equation \(0=-3(x+2)^{2}+9\), we get \(x = \pm \sqrt{3} - 2\). However, remember that \(f(x)=-3(x+2)^{2}+9\) is a downwards-opening parabola (because the leading coefficient is negative), therefore in this case, the parabola does not cross the x-axis and thus does not have any x-intercept.

Step 4: Plot the Function

Locate the vertex on the graph at the point (-2, 9). Then, plot the y-intercept at the point (0, -3). As there are no x-intercepts, the graph does not cross the x-axis. The parabola opens downwards due to the negative leading coefficient. Sketch the parabola by using the symmetry property with respect to the y-intercept and vertex, making sure the parabola is narrow due to the magnitude of the leading coefficient being greater than 1.

Key Concepts

Vertex Form

The vertex form of a quadratic function is a very useful way to understand the shape and position of a parabola. The general form is given by: \[ f(x) = a(x-h)^2 + k \] Where: - \( a \) determines the width and the direction (upwards or downwards) of the parabola. If \( a > 0 \), the parabola opens upwards; if \( a < 0 \), it opens downwards. - \( (h, k) \) is the vertex of the parabola, which is the highest or lowest point of the graph depending on the direction of the opening. In our function \( f(x) = -3(x+2)^2 + 9 \), we can see that \( a = -3 \), which suggests that the parabola opens downwards. The vertex coordinates are \( (h, k) = (-2, 9) \). Understanding the vertex form allows you to easily identify the vertex and how the graph will look without needing to calculate further points.

Parabola Properties

Parabolas have distinct properties that define their geometric shape on a graph. Let's break down these properties for a better understanding: **Direction of Opening**: - The sign of \( a \) in the vertex form \( f(x) = a(x-h)^2 + k \) tells you whether the parabola opens upward or downward. Here, since \( a = -3 \), the parabola opens downwards.
**Width of the Parabola**: - The absolute value of \( a \) impacts the width or narrowness of the parabola. With \( |a| = 3 \), our parabola is relatively narrow, indicating that it is steeper than if \( |a| = 1 \).
**Vertex as a Key Point**:- The vertex \((-2, 9)\) is a crucial point and acts as the maximum for our downward-opening parabola. It is the peak of the graph because the parabola does not continue upwards. By understanding these properties, you can predict how a parabola will behave and position itself on a graph.

Intercepts of Quadratic Functions

Intercepts are important points that describe where the parabola crosses the axes of a graph. **Y-Intercept**: - To find where the parabola crosses the y-axis, set \( x = 0 \) in the quadratic equation and solve for \( y \). For our function \( f(x) = -3(x+2)^2 + 9 \), we substitute \( x = 0 \): \[ f(0) = -3(0+2)^2 + 9 = -3(4) + 9 = -3 \] Thus, the y-intercept is at the point \((0, -3)\). **X-Intercepts**: - X-intercepts occur where \( f(x) = 0 \). Solving \( 0 = -3(x+2)^2 + 9 \), we find \[ (x+2)^2 = 3 \] Solving gives \( x = -2 \pm \sqrt{3} \). However, the negative leading coefficient confirms that the parabola does not actually cross the x-axis in this exercise, as the vertex is above the x-axis and the parabola opens downwards. Therefore, no real x-intercepts exist for this graphic representation. Understanding intercepts provides critical points that help define the shape and position of your quadratic graph in the Cartesian plane.