Problem 43 - 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. $$ f(x)=\frac{1}{2} x^{2}+3 x+\frac{9}{2} $$

Step-by-Step Solution

Answer

The y-intercept is \( \frac{9}{2} \), axis of symmetry is \( x = -3 \), and the vertex is \( (-3, 0) \). Graph the function using these points.

Step 1: Identify the Given Quadratic Function

We have the quadratic function: \( f(x) = \frac{1}{2}x^2 + 3x + \frac{9}{2} \). This is in the standard form \( ax^2 + bx + c \) where \( a = \frac{1}{2} \), \( b = 3 \), and \( c = \frac{9}{2} \).

Step 2: Find the y-intercept

The \( y \)-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the function: \( f(0) = \frac{1}{2}(0)^2 + 3(0) + \frac{9}{2} = \frac{9}{2} \). So, the \( y \)-intercept is \( \frac{9}{2} \).

Step 3: Find the Axis of Symmetry

The equation for the axis of symmetry of a quadratic function \( ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \). Here, \( b = 3 \) and \( a = \frac{1}{2} \), so:\[ x = -\frac{3}{2 \times \frac{1}{2}} = -3 \]Thus, the axis of symmetry is \( x = -3 \).

Step 4: Find the x-coordinate of the Vertex

The \( x \)-coordinate of the vertex is the same as the axis of symmetry, which is \( x = -3 \).

Step 5: Find the y-coordinate of the Vertex

To find the \( y \)-coordinate of the vertex, plug \( x = -3 \) into the function:\[ f(-3) = \frac{1}{2}(-3)^2 + 3(-3) + \frac{9}{2} \]\[ = \frac{1}{2}(9) - 9 + \frac{9}{2} \]\[ = \frac{9}{2} - 9 + \frac{9}{2} \]\[ = \frac{9}{2} + \frac{9}{2} - 9 \]\[ = 9 - 9 = 0 \]Thus, the vertex is \((-3, 0)\).

Step 6: Create a Table of Values

Select values of \( x \) around the vertex \( x = -3 \) to plot the graph. Here is a sample table of values:\[\begin{array}{c|c} x & f(x) \ \hline -5 & \frac{1}{2}(-5)^2 + 3(-5) + \frac{9}{2} = \frac{19}{2} \ -4 & \frac{1}{2}(-4)^2 + 3(-4) + \frac{9}{2} = 1 \ -3 & 0 \ -2 & \frac{1}{2}(-2)^2 + 3(-2) + \frac{9}{2} = \frac{5}{2} \ -1 & \frac{1}{2}(-1)^2 + 3(-1) + \frac{9}{2} = 4 \\end{array}\]

Step 7: Graph the Quadratic Function

Using the table of values and the key points: the \( y \)-intercept (\( 0, \frac{9}{2} \)), the vertex (\( -3, 0 \)), and symmetry about \( x = -3 \), plot the points on a coordinate plane. Then, draw a parabolic curve through these points to represent the function. The parabola opens upward since \( a = \frac{1}{2} > 0 \).

Key Concepts

Vertex of a Quadratic

The vertex of a quadratic function is a pivotal element in understanding its geometry. A vertex represents the highest or lowest point on the graph of a parabola, depending on whether the parabola opens upwards or downwards. In this instance, the quadratic is given by the function \( f(x) = \frac{1}{2}x^2 + 3x + \frac{9}{2} \). The vertex can be found using the formula for the vertex

• \

Axis of Symmetry

The axis of symmetry is a fundamental part of understanding quadratics. It is a vertical line that divides the parabola into two mirror-image halves. Think of it as the parabola's "balancing line." For a quadratic in the standard form \( ax^2 + bx + c \), the equation for the axis of symmetry is given by \( x = -\frac{b}{2a} \).
For our quadratic \( f(x) = \frac{1}{2}x^2 + 3x + \frac{9}{2} \), we use:

• \( a = \frac{1}{2} \)
• \( b = 3 \)

By substituting these values into the formula, \( x = -\frac{3}{2 \times \frac{1}{2}} = -3 \), we determine that the axis of symmetry is located at \( x = -3 \).
This axis is crucial when plotting the quadratic, as the vertex lies precisely on it, guiding the direction of the parabola's opening.

Y-intercept

The \( y \)-intercept of a quadratic function is where the graph crosses the \( y \)-axis of the coordinate plane. Finding the \( y \)-intercept provides a starting point for plotting the graph of the function.
For our quadratic function \( f(x) = \frac{1}{2}x^2 + 3x + \frac{9}{2} \), the \( y \)-intercept occurs when \( x = 0 \). Substituting \( x=0 \) into the equation:

• \( f(0) = \frac{1}{2}(0)^2 + 3(0) + \frac{9}{2} = \frac{9}{2} \).

Thus, the \( y \)-intercept is \( \frac{9}{2} \). This point indicates where the parabola intersects the \( y \)-axis and is an essential feature while drawing the graph.

Parabola

A parabola is the U-shaped curve that is depicted by a quadratic function. The direction of this curve is determined by the coefficient of \( x^2 \). In our example function \( f(x) = \frac{1}{2}x^2 + 3x + \frac{9}{2} \),

• The leading coefficient is \( \frac{1}{2} \), which is positive.

This means the parabola opens upwards. Features of a parabola include:

• The vertex, which can be the minimum or maximum point depending on the parabola's direction (here it is a minimum).
• The axis of symmetry, creating a mirror image from the vertex.
• Intersections with the axes, which help plot the function.

Understanding the parabola is vital as it helps visualize the nature and behavior of quadratic functions.

Table of Values

A table of values is a useful tool in graphing quadratic functions. It allows us to find several points on the parabola, especially around the vertex, to plot on the coordinate plane. This visual representation aids in understanding the shape and position of the quadratic graph.
Given the function \( f(x) = \frac{1}{2}x^2 + 3x + \frac{9}{2} \), we create a table of values by choosing \( x \)-values around the vertex cord \( x = -3 \). By substituting these \( x \)-values into the function, we acquire various \( (x,f(x)) \) points:

• For \( x = -5 \), \( f(x) = \frac{19}{2} \).
• For \( x = -4 \), \( f(x) = 1 \).
• For \( x = -3 \), \( f(x) = 0 \).
• For \( x = -2 \), \( f(x) = \frac{5}{2} \).
• For \( x = -1 \), \( f(x) = 4 \).

These points form the framework for sketching the parabola and help identify the curvature and intercepts visually.