Chapter 12: Problem 45
Graph each function using the vertex formula. Include the intercepts. \(f(x)=-x^{2}-8 x-13\)
Short Answer
Expert verified
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 by step solution
01
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 \((-4, 3)\).
02
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 \((0,-13)\).
03
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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,
For the exercise function \( f(x) = -x^2 - 8x - 13 \), we identified \( a = -1 \) and \( b = -8 \).
This point indicates the maximum point of the parabola since the parabola opens downward.
- \( f(x) = ax^2 + bx + c \)
- \( x = \frac{-b}{2a} \)
For the exercise function \( f(x) = -x^2 - 8x - 13 \), we identified \( a = -1 \) and \( b = -8 \).
- Using the formula:
- \( x_{vertex} = \frac{-(-8)}{2(-1)} = -4 \)
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:
\( -x^2 - 8x - 13 = 0 \), using the values of \( a = -1, b = -8, \text{ and } c = -13 \):
Plugging into the quadratic formula gives:
The x-intercepts are at
- \( f(x) = 0 \)
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( -x^2 - 8x - 13 = 0 \), using the values of \( a = -1, b = -8, \text{ and } c = -13 \):
- \( x = \frac{8 \pm \sqrt{64 - 52}}{-2} \)
- \( x = \frac{8 \pm \sqrt{12}}{-2} \)
The x-intercepts are at
- \( x = \frac{8 - \sqrt{12}}{-2} \)
- \( x = \frac{8 + \sqrt{12}}{-2} \)
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
To find this specific point, simply substitute \( x = 0 \) into the quadratic function and solve for \( f(x) \). It gives you the point
- \( (0, c) \)
- By setting \( x = 0 \):
- \( f(0) = -(0)^2 - 8 \, \times \, 0 - 13 = -13 \)
- \( (0, -13) \)
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
- \( ax^2 + bx + c = 0 \)
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- \( b^2 - 4ac \)
- 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).
- \( b^2 - 4ac = 64 - 52 = 12 \)