Problem 35 - College Algebra

Question

For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts. $$ f(x)=x^{2}-6 x-1 $$

Step-by-Step Solution

Answer

Vertex: (3, -10); Axis of Symmetry: x = 3; Intercepts: (0, -1), (3+√10, 0), (3-√10, 0).

Step 1: Identify the Form of the Quadratic Equation

The given quadratic function is in the standard form \( f(x) = ax^2 + bx + c \) where \( a = 1 \), \( b = -6 \), and \( c = -1 \). This helps in identifying the parameters to find the vertex and axis of symmetry.

Step 2: Calculate the Vertex

The vertex \( (h, k) \) of a parabola in standard form is given by \( h = -\frac{b}{2a} \) and \( k = f(h) \). Using the values from the equation, \( h = -\frac{-6}{2 \cdot 1} = 3 \). Substitute \( h \) back into the function to find \( k \):\[ k = f(3) = (3)^2 - 6(3) - 1 = 9 - 18 - 1 = -10. \]So, the vertex of the parabola is \( (3, -10) \).

Step 3: Determine the Axis of Symmetry

The axis of symmetry of a parabola in standard form is a vertical line \( x = h \). From Step 2, we found \( h = 3 \). Therefore, the axis of symmetry is \( x = 3 \).

Step 4: Find the Intercepts

- **Y-intercept**: This occurs where \( x = 0 \). Substitute \( x = 0 \) into the function: \( f(0) = 0^2 - 6(0) - 1 = -1 \). So, the y-intercept is \( (0, -1) \).- **X-intercepts**: These occur where \( f(x) = 0 \). Solve \( x^2 - 6x - 1 = 0 \) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}. \]Calculate:\[ x = \frac{6 \pm \sqrt{36 + 4}}{2} = \frac{6 \pm \sqrt{40}}{2} = \frac{6 \pm 2\sqrt{10}}{2} = 3 \pm \sqrt{10}. \]Hence, the x-intercepts are \( (3 + \sqrt{10}, 0) \) and \( (3 - \sqrt{10}, 0) \).

Step 5: Sketch the Graph

Plot the vertex \( (3, -10) \), the axis of symmetry \( x = 3 \), the y-intercept \( (0, -1) \), and the x-intercepts \( (3 + \sqrt{10}, 0) \) and \( (3 - \sqrt{10}, 0) \) on a coordinate plane. Then draw a smooth curve through these points, ensuring the parabola opens upwards, as \( a = 1 > 0 \).

Key Concepts

Vertex of a Parabola

The vertex of a parabola is the point where the curve reaches its maximum or minimum value. It's a crucial component in understanding the graph of a quadratic function. The vertex can be found with the formula for a quadratic equation in standard form \( f(x) = ax^2 + bx + c \).
For our quadratic \( f(x) = x^2 - 6x - 1 \), the vertex \((h, k)\) is calculated where \( h = -\frac{b}{2a} \). Substituting in the values gives us \( h = 3 \).
The y-value \( k \) at this point is found by evaluating the function at \( x = 3 \):\[ k = 3^2 - 6 \times 3 - 1 = -10. \]Thus, the vertex is \((3, -10)\). This point acts as the turning point of the parabola. If \( a > 0 \), as it is here, the parabola opens upwards, making the vertex the lowest point on the graph.

Axis of Symmetry

The axis of symmetry is an essential feature that divides the parabola into two mirror-image halves. It is a vertical line that passes through the vertex of the parabola.
For quadratic functions in standard form, this line can be defined as \( x = h \), where \( h \) is the x-coordinate of the vertex. Using the example \( f(x) = x^2 - 6x - 1 \), we found \( h = 3 \).
Hence, the axis of symmetry is the line \( x = 3 \).

This line helps to visualize the parabola and make predictions about the graph's shape and direction. On either side of this line, the y-values of the parabola are equidistant, reinforcing the symmetrical nature of quadratic graphs.

Intercepts of a Quadratic

Intercepts are the points at which the graph of a quadratic function crosses the axes. There are two main types of intercepts:

• Y-intercept: This occurs where the graph crosses the y-axis (where \( x = 0 \)). By substituting \( x = 0 \) in \( f(x) = x^2 - 6x - 1 \), we find the y-intercept is \((0, -1)\).
• X-intercepts: These are the points where the graph crosses the x-axis (where \( f(x) = 0 \)). To find these, solve \( x^2 - 6x - 1 = 0 \) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]which results in \( x = 3 \pm \sqrt{10} \). This gives us two x-intercepts at \( (3 + \sqrt{10}, 0) \) and \( (3 - \sqrt{10}, 0) \).

Intercepts are integral in shaping the parabola and are key points to plot when graphing quadratic functions.

Graphing Parabolas

Graphing parabolas requires plotting points such as the vertex, intercepts, and understanding the parabola's symmetry. Begin by plotting the vertex, here \((3, -10)\), the lowest point of our upward-opening parabola.

Next, draw the axis of symmetry \(x = 3\), which guides the symmetry of the graph.

• Plot the y-intercept \((0, -1)\) - this is where the parabola crosses the y-axis.
• Mark the x-intercepts \((3 + \sqrt{10}, 0)\) and \((3 - \sqrt{10}, 0)\) on the graph.
0 \).
Graphing provides a visual understanding of how quadratic functions behave and clearly shows the important features such as vertex, axis of symmetry, and intercepts.