Problem 44 - 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. $$ f(x)=x^{2}-\frac{2}{3} x-\frac{8}{9} $$

Step-by-Step Solution

Answer

The y-intercept is \((0, -\frac{8}{9})\), the axis is \(x = \frac{1}{3}\), and the vertex is \((\frac{1}{3}, -1)\). Graph using these points.

Step 1: Find the y-intercept

To find the \( y \)-intercept, we set \( x = 0 \). Substitute \( x = 0 \) into the function: \[f(0) = (0)^2 - \frac{2}{3}(0) - \frac{8}{9} = -\frac{8}{9}.\]Thus, the \( y \)-intercept is \((0, -\frac{8}{9})\).

Step 2: Find the Axis of Symmetry and Vertex X-coordinate

For a quadratic function \( f(x) = ax^2 + bx + c \), the equation of the axis of symmetry is \( x = -\frac{b}{2a} \).Here, \( a = 1 \) and \( b = -\frac{2}{3} \).Calculate: \[x = -\frac{-\frac{2}{3}}{2 \times 1} = \frac{1}{3}.\]So, the axis of symmetry is \( x = \frac{1}{3} \) and the \( x \)-coordinate of the vertex is \( \frac{1}{3} \).

Step 3: Calculate the Vertex Y-coordinate

To find the \( y \)-coordinate of the vertex, substitute the \( x \)-coordinate of the vertex back into the function: \[f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^2 - \frac{2}{3}\times\frac{1}{3} - \frac{8}{9}.\]Calculate:\[f\left(\frac{1}{3}\right) = \frac{1}{9} - \frac{2}{9} - \frac{8}{9} = -\frac{9}{9} = -1.\]Therefore, the vertex is at \( \left(\frac{1}{3}, -1\right) \).

Step 4: Make a Table of Values Including the Vertex

Create a table with selected values around the vertex \( x = \frac{1}{3} \). Calculate corresponding \( y \)-values for \( x = 0 \), \( x = 1 \), and include the vertex \( x = \frac{1}{3} \):- \( x = 0 \), \( y = -\frac{8}{9} \). - \( x = \frac{1}{3} \), \( y = -1 \) (vertex).- \( x = 1 \), \( y = f(1) = 1 - \frac{2}{3} - \frac{8}{9} = -\frac{11}{9} \).| \(x\) | \(y\) ||-------|--------------|| 0 | -\(\frac{8}{9}\) || \(\frac{1}{3}\) | -1 || 1 | -\(\frac{11}{9}\) |

Step 5: Graph the Function

Now we use the intercepts and vertex identified, as well as the table of values, to plot the function on a coordinate grid. Plot the points (0, -8/9), (1/3, -1), and (1, -11/9), and draw a symmetric parabola opening upwards. The axis of symmetry is the vertical line \( x = \frac{1}{3} \).

Key Concepts

Understanding the y-intercept in Quadratic Functions

The y-intercept for a quadratic function is the point where the graph meets the y-axis. It's essentially where the function "starts" in terms of its height on the graph. To find the y-intercept, you set x to zero and evaluate the function at that point. For example, in the quadratic function we have, plugging in x = 0 gives us:\[f(0) = 0^2 - \frac{2}{3} \cdot 0 - \frac{8}{9} = -\frac{8}{9}.\]

This results in the y-intercept being at the coordinate \((0, -\frac{8}{9})\). This point is crucial, as it aids in plotting the initial position of the parabola on the graph, showing where it intersects the y-axis.

Defining the Axis of Symmetry in Quadratic Graphs

The axis of symmetry is a vital part of understanding parabolas, as it indicates the line that perfectly divides the parabola into two mirrored halves. To find this line in a quadratic function formatted as \(f(x) = ax^2 + bx + c\), you use the formula:\[x = -\frac{b}{2a}.\]

For our current function, \(a = 1\) and \(b = -\frac{2}{3}\), leading to:\[x = -\frac{-\frac{2}{3}}{2 \times 1} = \frac{1}{3}.\]

This defines the axis of symmetry as the line \(x = \frac{1}{3}\). This line is helpful in predicting the shape and spread of the parabola, acting like a spine along which the graph curves symmetrically.

Finding and Interpreting Vertex Coordinates

In quadratic functions, the vertex represents the peak or the lowest point, depending on the parabola's orientation. It's at this point where the curve changes its direction. The vertex is located on the axis of symmetry. For the given function, the x-coordinate of the vertex is \(\frac{1}{3}\). To find the y-coordinate, place this x-value back into the original function:\[f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^2 - \frac{2}{3} \cdot \frac{1}{3} - \frac{8}{9}.\]

This evaluates to:\[\frac{1}{9} - \frac{2}{9} - \frac{8}{9} = -1.\]

Thus, the vertex is at \(\left(\frac{1}{3}, -1\right)\). This point is essential as it reveals the lowest point on this upward-opening parabola.

Using a Table of Values for Accurate Graphing

A table of values is a handy tool when graphing quadratic functions. It provides specific points that lie on the parabola, making drawing the curve much easier. For the quadratic function, include critical points like the vertex, and other nearby x-values to illustrate the parabola's shape.

• For \(x = 0\), the y-value is \(-\frac{8}{9}.\)
• For the vertex \(x = \frac{1}{3}\), the y-value is \(-1.\)
• For \(x = 1\), compute \(f(1) = 1 - \frac{2}{3} - \frac{8}{9} = -\frac{11}{9}.\)

The values are:

- \((0, -\frac{8}{9})\)
- \((\frac{1}{3}, -1)\) (vertex)
- \((1, -\frac{11}{9})\)

Using these points helps plot a more precise graph, as it ensures the plotted curve accurately represents the quadratic function.