Problem 13 - Algebra 2

Question

Complete parts a-c for each quadratic function. a. Find the \(y\) -intercept, the equation of the axis of symmetry, and the \(x\) -coordinate of the vertex. b. Make a table of values that includes the vertex. c. Use this information to graph the function. $$ f(x)=-5 x^{2} $$

Step-by-Step Solution

Answer

The y-intercept is (0, 0), axis of symmetry is x = 0; graph shows parabola opening downward with vertex (0, 0).

Step 1: Determine the y-intercept

For a quadratic function in the form of \( f(x) = ax^2 + bx + c \), the \( y \)-intercept is the value of the function when \( x = 0 \). Here, \( f(x) = -5x^2 \), so when \( x = 0 \), \( f(0) = -5(0)^2 = 0 \). Thus, the \( y \)-intercept is \( (0, 0) \).

Step 2: Find the Equation of the Axis of Symmetry

The equation of the axis of symmetry for a quadratic function \( f(x) = ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \). For \( f(x) = -5x^2 \), we have \( a = -5 \) and \( b = 0 \), so the axis of symmetry is \( x = -\frac{0}{2(-5)} = 0 \).

Step 3: Determine the x-coordinate of the Vertex

In a quadratic function \( f(x) = ax^2 + bx + c \), the \( x \)-coordinate of the vertex is the same as the axis of symmetry. From Step 2, we found that the axis of symmetry is \( x = 0 \), meaning the \( x \)-coordinate of the vertex is also 0.

Step 4: Construct the Table of Values

To create a table of values for \( f(x) = -5x^2 \), we choose values of \( x \) around the vertex \( x = 0 \). A suitable table might include points \( x = -2, -1, 0, 1, 2 \). Calculating each: \[\begin{array}{c|c}x & f(x) \\hline-2 & -5(-2)^2 = -20 \-1 & -5(-1)^2 = -5 \0 & -5(0)^2 = 0 \1 & -5(1)^2 = -5 \2 & -5(2)^2 = -20 \\end{array}\]

Step 5: Graph the Function

Using the table of values from Step 4, plot the points \((-2, -20), (-1, -5), (0, 0), (1, -5), (2, -20)\) on a coordinate plane. Draw a smooth curve through these points forming a parabola opening downward. The vertex at \((0, 0)\) is the highest point since the parabola opens downward due to the negative coefficient of \( x^2 \).

Key Concepts

y-intercept

The concept of the y-intercept in quadratic functions, or any other type of function, is quite straightforward. It is the point where the graph of the function crosses the y-axis. For any function in the standard quadratic form of \( f(x) = ax^2 + bx + c \), the y-intercept is found by substituting \( x = 0 \) into the equation, resulting in the point \( (0, c) \).

In our example problem, the function \( f(x) = -5x^2 \) does not contain a \( bx \) term or a constant term \( c \), so it simplifies to \( f(0) = -5(0)^2 = 0 \). This means the y-intercept is at the origin point \( (0, 0) \).

When graphing functions, the y-intercept serves as a starting point, and it's a fixed location on the vertical axis that helps in sketching the overall shape of the graph, especially when combined with other features like the vertex and axis of symmetry.

axis of symmetry

The axis of symmetry is an important concept when dealing with quadratic functions. It is the vertical line that divides the parabola into two mirror images, essentially reflecting one side onto the other. This line passes through the vertex of the parabola.

For any quadratic function in the form \( f(x) = ax^2 + bx + c \), the equation of the axis of symmetry can be found using the formula \( x = -\frac{b}{2a} \).

In the given exercise for \( f(x) = -5x^2 \), the coefficients are \( a = -5 \) and \( b = 0 \). Substituting these into the formula, we get \( x = -\frac{0}{2(-5)} = 0 \).

This tells us that the axis of symmetry for this parabola is the vertical line \( x = 0 \). This line plays a crucial role when graphing the function because it helps to ensure the parabola is drawn symmetrically.

vertex

The vertex of a quadratic function is a significant point that provides a lot of information about the graph. It is the peak point of the parabola, being either the highest or lowest point depending on the parabola's orientation.

The vertex lies on the axis of symmetry. Thus, its \( x \)-coordinate is the same as the equation of the axis of symmetry: \( x = -\frac{b}{2a} \).

For \( f(x) = -5x^2 \), with \( a = -5 \) and \( b = 0 \), the vertex's \( x \)-coordinate is \( 0 \). The \( y \)-coordinate is found by substituting \( x = 0 \) back into the function, which computes as \( f(0) = -5(0)^2 = 0 \).

Hence, the vertex of the function is at the point \( (0, 0) \). In the context of graphing, the vertex serves as a turning point and is essential in shaping the curve of the parabola effectively.

graphing functions

Graphing a quadratic function involves several key steps and using different properties of the function.

To graph \( f(x) = -5x^2 \), we start by identifying important points:

• The y-intercept: This step gives us the point \( (0, 0) \).
• The vertex: From previous sections, we know the vertex is located at \( (0, 0) \).
• The axis of symmetry: As established, it is \( x = 0 \), guiding us in placing points symmetrically.

Next, construct a table of values by choosing several \( x \)-values around the vertex. Typical choices are \( x = -2, -1, 0, 1, 2 \). Calculations from the original exercise show the values for \( f(x) \):

• \( f(-2) = -20 \)
• \( f(-1) = -5 \)
• \( f(0) = 0 \)
• \( f(1) = -5 \)
• \( f(2) = -20 \)

Finally, plot these points on a coordinate plane and draw a smooth curve through them. The graph should form a downward-opening parabola due to the negative coefficient \( a = -5 \). The symmetry, y-intercept, and vertex help ensure the accuracy of the parabola's shape.