Problem 36 - College Algebra with Corequisite Support

Question

For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts. $$ f(x)=x^{2}-5 x-6 $$

Step-by-Step Solution

Answer

Vertex: \( \left( \frac{5}{2}, -\frac{1}{4} \right) \), Axis of symmetry: \( x = \frac{5}{2} \), Y-intercept: \( (0, -6) \), X-intercepts: \( (6, 0) \) and \( (-1, 0) \).

Step 1: Identify the form of the quadratic

The given quadratic function is \( f(x) = x^2 - 5x - 6 \). This is in the standard form \( ax^2 + bx + c \), where \( a = 1 \), \( b = -5 \), and \( c = -6 \).

Step 2: Find the vertex

The vertex of a quadratic in standard form \( ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \). For our function, \( a = 1 \) and \( b = -5 \), so \( x = -\frac{-5}{2 \times 1} = \frac{5}{2} \). Substitute \( x = \frac{5}{2} \) back into the function to find the y-coordinate: \[ f\left(\frac{5}{2}\right) = \left( \frac{5}{2} \right)^2 - 5\left( \frac{5}{2} \right) - 6 = \frac{25}{4} - \frac{25}{2} - 6 = -\frac{1}{4} \]. Thus, the vertex is \( \left( \frac{5}{2}, -\frac{1}{4} \right) \).

Step 3: Determine the axis of symmetry

The axis of symmetry is the vertical line that passes through the vertex of the parabola. From the vertex calculation, we know the axis of symmetry is the line \( x = \frac{5}{2} \).

Step 4: Find the y-intercept

The y-intercept is found by evaluating the function when \( x = 0 \). Substitute \( x = 0 \) into \( f(x) \): \[ f(0) = 0^2 - 5 \times 0 - 6 = -6 \]. Thus, the y-intercept is \( (0, -6) \).

Step 5: Find the x-intercepts

To find the x-intercepts, solve the equation \( x^2 - 5x - 6 = 0 \). This can be factored as \( (x - 6)(x + 1) = 0 \). Thus, the solutions are \( x = 6 \) and \( x = -1 \). Hence, the x-intercepts are \( (6, 0) \) and \( (-1, 0) \).

Step 6: Sketch the graph

Plot the vertex \( \left( \frac{5}{2}, -\frac{1}{4} \right) \), the axis of symmetry \( x = \frac{5}{2} \), the y-intercept \( (0, -6) \), and the x-intercepts \( (6, 0) \) and \( (-1, 0) \). Draw a smooth parabola opening upwards through these points.

Key Concepts

Vertex Calculation

When dealing with quadratic functions, finding the vertex is pivotal for understanding the shape and placement of the parabola on a graph. In the standard form of a quadratic equation, which is given by \( f(x) = ax^2 + bx + c \), the vertex's x-coordinate is calculated using the formula \( x = -\frac{b}{2a} \). This formula stems from completing the square, a technique used for transforming quadratic equations.Once you have the x-coordinate, substitute it back into the original quadratic equation to find the y-coordinate. For the function \( f(x) = x^2 - 5x - 6 \), the x-coordinate of the vertex is \( \frac{5}{2} \), and when plugged back into the equation, the corresponding y-value is \( -\frac{1}{4} \). Thus, the vertex is \( \left( \frac{5}{2}, -\frac{1}{4} \right) \).

• The vertex provides the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards.
• In this example, because the coefficient of \( x^2 \) is positive, the parabola opens upwards, making the vertex a minimum point.

Axis of Symmetry

The axis of symmetry of a parabola is a crucial line that divides it into two mirror-image halves. For any quadratic in the form \( f(x) = ax^2 + bx + c \), the axis of symmetry can be found using the vertex's x-coordinate, specifically at \( x = -\frac{b}{2a} \).In simple terms, this line runs vertically through the vertex. Knowing this helps in drawing the parabola accurately as every point on one side of the axis has a corresponding point on the opposite side.

• In the quadratic \( f(x) = x^2 - 5x - 6 \), the axis of symmetry is the vertical line \( x = \frac{5}{2} \).
• This line isn't part of the graph itself but guides the reflection symmetry of the parabola.

Intercepts of a Parabola

Intercepts are the points where the parabola crosses the x-axis and y-axis, which are essential for understanding the function's behavior in different regions.

Y-Intercept:

The y-intercept is found by setting \( x = 0 \) in the quadratic equation. For \( f(x) = x^2 - 5x - 6 \), substituting \( x = 0 \) gives \( f(0) = -6 \), so the y-intercept is at the point \( (0, -6) \).

X-Intercepts:

X-intercepts occur where the function crosses the x-axis (where \( f(x) = 0 \)). Solving \( x^2 - 5x - 6 = 0 \) by factoring yields two solutions: \( x - 6 = 0 \) giving \( x = 6 \), and \( x + 1 = 0 \) giving \( x = -1 \).

• The x-intercepts are the points \( (6, 0) \) and \( (-1, 0) \), marking where the parabola touches the x-axis.
• These intercepts help in understanding the width and location of the parabola on a graph.