Problem 52 - Algebra 2

Question

Find the \(y\) -intercept, the equation of the axis of symmetry, and the \(x-\) -coordinate of the vertex for each quadratic function. Then graph the function by making a table of values. $$ f(x)=x^{2}-6 x+4 $$

Step-by-Step Solution

Answer

The y-intercept is 4, the axis of symmetry is \(x = 3\), and the vertex has an x-coordinate of 3. Plot the points from the table for the graph.

Step 1: Identify the Quadratic Equation Form

The given quadratic equation is in the form of \( f(x) = ax^2 + bx + c \). Here, \( a = 1 \), \( b = -6 \), and \( c = 4 \).

Step 2: Find the Y-Intercept

The \( y \)-intercept of the quadratic function can be found by setting \( x = 0 \). Plugging \( x = 0 \) into the equation gives \( f(0) = 0^2 - 6(0) + 4 = 4 \). Thus, the \( y \)-intercept is \( (0, 4) \).

Step 3: Determine the Axis of Symmetry

The axis of symmetry for a parabola given by \( f(x) = ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = -6 \), so \( x = -\frac{-6}{2(1)} = 3 \). 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 value for the axis of symmetry, which is \( x = 3 \).

Step 5: Create a Table of Values to Graph the Function

Choose values around the vertex to find corresponding \( y \) values. The table should include \( x = 0, 1, 2, 3, 4, 5, 6 \):- For \( x = 0 \), \( f(0) = 4 \)- For \( x = 1 \), \( f(1) = 1^2 - 6(1) + 4 = -1 \)- For \( x = 2 \), \( f(2) = 2^2 - 6(2) + 4 = -4 \)- For \( x = 3 \), \( f(3) = 3^2 - 6(3) + 4 = -5 \) (vertex)- For \( x = 4 \), \( f(4) = 4^2 - 6(4) + 4 = -4 \)- For \( x = 5 \), \( f(5) = 5^2 - 6(5) + 4 = -1 \)- For \( x = 6 \), \( f(6) = 6^2 - 6(6) + 4 = 4 \)

Step 6: Graph the Function Using the Table

Use the table of values to plot points on a graph. The curve will be symmetrical around the line \( x = 3 \), which is the axis of symmetry, and will pass through the points generated in the table. The vertex of the parabola is at \( (3, -5) \).

Key Concepts

Y-Intercept

The y-intercept of a quadratic function is the point where the graph crosses the y-axis. To find this point, we set the value of \( x = 0 \). This simplifies our equation to just the constant term, since the other terms will multiply zero and vanish.
In our function \( f(x) = x^2 - 6x + 4 \), we substitute \( x = 0 \) which results in \( f(0) = 0^2 - 6(0) + 4 = 4 \).
Thus, the y-intercept is at the point \( (0, 4) \).

• This point is helpful in graphing because it gives us a starting point on the vertical axis.
• Regardless of what the other coefficients are, the y-intercept only depends on the constant in the equation.

Axis of Symmetry

The axis of symmetry of a parabola is a vertical line that divides it into two mirror-image halves. For any quadratic function \( f(x) = ax^2 + bx + c \), this line is given by the formula \( x = -\frac{b}{2a} \).
In our case:

• \( a = 1 \)
• \( b = -6 \)

Plugging these values into the formula, we get:\[ x = -\frac{-6}{2(1)} = 3 \]Therefore, the axis of symmetry is \( x = 3 \).
This line is crucial because:

• The parabola is symmetrical about this line.
• The x-coordinate of the vertex is the same as the axis of symmetry.

Vertex of a Parabola

The vertex of a parabola is the highest or lowest point on the graph, depending on its orientation. For a standard quadratic function, the x-coordinate of the vertex is the same as the axis of symmetry. That means it is found at \( x = -\frac{b}{2a} \).
Given our equation, we've already found that this occurs at \( x = 3 \).
To find the complete vertex (both x and y coordinates), we substitute \( x = 3 \) back into our function:\[ f(3) = 3^2 - 6(3) + 4 = -5 \]Thus, the vertex of the parabola is at the point \( (3, -5) \).

• This point represents the minimum value of the parabola because the coefficient \( a = 1 \) is positive, indicating the parabola opens upwards.
• The vertex is a critical point in graphing, as it helps determine the parabola's "bowl" shape.

Graphing a Parabola

Graphing a parabola involves plotting points on a coordinate plane and joining them in a curved shape. Utilizing the vertex, y-intercept, and axis of symmetry provides a guide for this process.
First, make a table of values. In our example, we chose values of \( x \) around the vertex:
\[ x = 0, 1, 2, 3, 4, 5, 6 \]Calculate the corresponding y-values:

• \( f(0) = 4 \)
• \( f(1) = -1 \)
• \( f(2) = -4 \)
• \( f(3) = -5 \) (vertex)
• \( f(4) = -4 \)
• \( f(5) = -1 \)
• \( f(6) = 4 \)

Now, plot these points and draw a smooth, U-shaped curve through them, ensuring symmetry around the axis.
Graphing tips:

• The axis of symmetry \( x = 3 \) helps to verify the balance of the graph.
• The vertex \( (3, -5) \) guides the peak or lowest point of the parabola.
• The y-intercept \( (0, 4) \) provides an initial point on the y-axis.