Problem 45 - Beginning and Intermediate Algebra

Question

Graph each function using the vertex formula. Include the intercepts. \(f(x)=-x^{2}-8 x-13\)

Step-by-Step Solution

Answer

The vertex of the parabola is \((-4, 3)\). The x-intercepts are at \(x = \frac{8 - \sqrt{12}}{-2}\) and \(x = \frac{8 + \sqrt{12}}{-2}\). The y-intercept is at \((0, -13)\). The parabola opens downward. Graph the function with these details marked.

Step 1: Identify the Vertex of the Parabola

The vertex of a quadratic function in the general form \(f(x) = ax^2 + bx + c\) is given by the formula: \[x = \frac{-b}{2a}\] In this function, we have \(a = -1, b = -8\), and \(c = -13\). Using the formula above, we can find the x-coordinate of the vertex: \[x_{vertex} = \frac{-(-8)}{2(-1)} = \frac{8}{-2} = -4\] Now that we have the x-coordinate of the vertex, we can find the corresponding y-coordinate by plugging the x-coordinate into the given function: \[f(-4) = -(-4)^2 - 8(-4) - 13 = -16 + 32 - 13 = 3\] The vertex of the parabola is at the point \((-4, 3)\).

Step 2: Find the x-intercepts and y-intercept

To find the x-intercepts, we need to find where the function equals zero. By solving the equation \(f(x) = 0\), we can determine the points at which the parabola intersects the x-axis. \(0 = -x^2 - 8x - 13\) As this quadratic equation does not factor easily, let's use the quadratic formula to solve for x: \[x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}\] For this function, we have: \[x = \frac{-(-8) \pm \sqrt{(-8)^2 - (4)(-1)(-13)}}{(2)(-1)}\] \[x = \frac{8 \pm \sqrt{64 - 52}}{-2}\] \[x = \frac{8 \pm \sqrt{12}}{-2}\] Thus, the x-intercepts are at \(x = \frac{8 - \sqrt{12}}{-2}\) and \(x = \frac{8 + \sqrt{12}}{-2}\). To find the y-intercept, plug in x = 0, and solve for f(x): \[f(0) = -(0)^2 - 8(0) - 13 = -13\] The y-intercept is at the point \((0,-13)\).

Step 3: Sketch the Graph

Now, using the information we gathered, we can sketch the graph: 1. The vertex is at \((-4, 3)\). 2. The x-intercepts are at \(x = \frac{8 - \sqrt{12}}{-2}\) and \(x = \frac{8 + \sqrt{12}}{-2}\) 3. The y-intercept is at \((0, -13)\) 4. The parabola opens downward (since a = -1) With these details, you can now graph the function \(f(x) = -x^2 - 8x - 13\). The vertex, x-intercepts, and y-intercept should be clearly marked on your graph.

Key Concepts

Vertex Formula

To find the vertex of a quadratic function, we rely on the vertex formula, which is especially useful when dealing with a parabolic equation in the standard form,

• \( f(x) = ax^2 + bx + c \)

The formula to determine the x-coordinate of the vertex is given by:

• \( x = \frac{-b}{2a} \)

In this equation, \( b \) and \( a \) are the coefficients of the linear and quadratic terms respectively. By plugging these values into the formula, you can find precisely where the vertex of the parabola sits along the x-axis. The y-coordinate is then found by substituting this x value back into the original function.

For the exercise function \( f(x) = -x^2 - 8x - 13 \), we identified \( a = -1 \) and \( b = -8 \).

• Using the formula:
  • \( x_{vertex} = \frac{-(-8)}{2(-1)} = -4 \)

So, the x-coordinate of the vertex is -4. By substituting back into the function, \( f(-4) = 3 \), revealing that the vertex is at the point \((-4, 3)\).
This point indicates the maximum point of the parabola since the parabola opens downward.

X-Intercepts

The x-intercepts of a quadratic function are the points where the graph crosses the x-axis. Mathematically, these are the solutions to the equation when the function is set to zero:

• \( f(x) = 0 \)

To find these intercepts for quadratic equations that cannot be easily factored, we can utilize the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula helps in finding the roots of any quadratic equation, offering solutions as values of x where the parabola touches or crosses the x-axis. For this exercise:
\( -x^2 - 8x - 13 = 0 \), using the values of \( a = -1, b = -8, \text{ and } c = -13 \):

Plugging into the quadratic formula gives:

• \( x = \frac{8 \pm \sqrt{64 - 52}}{-2} \)
• \( x = \frac{8 \pm \sqrt{12}}{-2} \)

The x-intercepts are at

• \( x = \frac{8 - \sqrt{12}}{-2} \)
• \( x = \frac{8 + \sqrt{12}}{-2} \)

These points are crucial for understanding where the quadratic will touch or cross the horizontal axis.

Y-Intercept

The y-intercept of a quadratic function is the point where the graph intersects the y-axis. This occurs when the x-coordinate is zero.
To find this specific point, simply substitute \( x = 0 \) into the quadratic function and solve for \( f(x) \). It gives you the point

• \( (0, c) \)

where \( c \) is the constant term in the function. Let's apply this to our example function, \( f(x) = -x^2 - 8x - 13 \):

• By setting \( x = 0 \):
  • \( f(0) = -(0)^2 - 8 \, \times \, 0 - 13 = -13 \)

This tells us that the y-intercept is at the point

• \( (0, -13) \)

This point indicates where the parabola crosses the y-axis on the Cartesian plane.

Quadratic Formula

The quadratic formula is a vital mathematical tool for solving any quadratic equation, especially when the equation isn’t easily factored. Any equation that takes on the quadratic form

• \( ax^2 + bx + c = 0 \)

can be addressed using this robust formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

It provides the x-values at which the quadratic graph intersects the x-axis, known as the roots or x-intercepts. The expression under the square root, known as the discriminant

• \( b^2 - 4ac \)

can tell us about the nature of these roots:

• If positive, two distinct real roots exist.
• If zero, there’s exactly one real root (the vertex touches the x-axis).
• If negative, no real roots exist (the parabola does not cross the x-axis).

In our example equation \( -x^2 - 8x - 13 = 0 \), after calculating

• \( b^2 - 4ac = 64 - 52 = 12 \)

We found two real roots, suggesting the parabola intersects with the x-axis at two points. This detailed insight provided by the quadratic formula is essential in precisely graphing quadratic functions and understanding their behaviors.