Chapter 8: Problem 19
(Graphing program required.) Given \(f(x)=-x^{2}+8 x-15\) : a. Estimate by graphing: the \(x\) -intercepts, the \(y\) -intercept, and the vertex. b. Calculate the coordinates of the vertex.
Short Answer
Expert verified
x-intercepts: estimated from graph, y-intercept: -15, vertex: (4, 1), vertex coordinates: (4, 1)
Step by step solution
01
- Graph the Function
Plot the function given by the equation \(f(x) = -x^2 + 8x - 15\) using a graphing program or graphing calculator. Observe the shape of the parabola.
02
- Identify the x-intercepts
Locate the points where the graph intersects the x-axis. These points are the x-intercepts. Estimate these points from the graph.
03
- Identify the y-intercept
Identify the point where the graph intersects the y-axis. This point is the y-intercept.
04
- Estimate the Vertex
Find the highest point on the graph of the parabola since the parabola opens downwards. This point is the vertex. Estimate the coordinates from the graph.
05
- Calculate the Vertex Coordinates
The vertex of a parabola given by the equation \(y = ax^2 + bx + c\) can be found using the formula \(x = \frac{-b}{2a}\). Substitute \(a = -1\), \(b = 8\), and \(c = -15\) to find the x-coordinate: \(x = \frac{-8}{2(-1)} = 4\). Substitute \(x = 4\) back into the equation to find the y-coordinate: \(y = -4^2 + 8(4) - 15 = -16 + 32 - 15 = 1\). So, the vertex is at \((4, 1)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
x-intercepts
In any quadratic function of the form \( f(x) = ax^2 + bx + c \), the x-intercepts are the points where the graph crosses the x-axis. These are the values of \( x \) for which \( f(x) = 0 \). For the equation \( f(x) = -x^2 + 8x - 15 \), solving for the x-intercepts involves finding the roots of the quadratic equation.
- First, set the equation equal to zero: \( -x^2 + 8x - 15 = 0 \).
- Next, use the quadratic formula: \( x = \frac{-b \, \text{\textpm} \, \text{\textsqrt}{b^2 - 4ac}}{2a} \), where \( a = -1 \), \( b = 8 \), and \( c = -15 \).
- Calculate the discriminant: \( b^2 - 4ac \). Here, it is \( 8^2 - 4(-1)(-15) = 64 - 60 = 4 \).
- Finally, find the x-intercepts: \( x = \frac{-8 \, \text{\textpm} \, \text{\textsqrt}{4}}{2(-1)} \). The solutions are \( x = 3 \) and \( x = 5 \).
y-intercept
The y-intercept of a quadratic function is the point where the graph crosses the y-axis. This occurs when \( x = 0 \). To find the y-intercept for the equation \( f(x) = -x^2 + 8x - 15 \), simply substitute \( x = 0 \) into the equation.
- Set \( x = 0 \) in the equation: \( f(0) = -0^2 + 8(0) - 15 \).
- Simplify the equation: \( f(0) = -15 \).
vertex calculation
The vertex of a parabola described by the quadratic function \( f(x) = ax^2 + bx + c \) represents its highest or lowest point, depending on the direction it opens. For the function \( f(x) = -x^2 + 8x - 15 \), the vertex can be found using the vertex formula, \( x = \frac{-b}{2a} \).
- Identify the coefficients: \( a = -1 \) and \( b = 8 \).
- Calculate the x-coordinate of the vertex: \( x = \frac{-8}{2(-1)} = 4 \).
- Substitute \( x = 4 \) in \( f(x) \): \( f(4) = -4^2 + 8(4) - 15 \).
- Simplify the expression: \( f(4) = -16 + 32 - 15 = 1 \).
graphing quadratic functions
Graphing a quadratic function involves understanding the shape and key features of its parabola. The general form is \( f(x) = ax^2 + bx + c \). Here's how to graph the function accurately:
- Identify the direction: If \( a \) is positive, the parabola opens upwards. If \( a \) is negative, it opens downwards. For \( f(x) = -x^2 + 8x - 15 \), \( a = -1 \), so the parabola opens downwards.
- Locate the key points: These include the x-intercepts, y-intercept, and vertex. Use the calculations described earlier. For our function, we have:
- x-intercepts: \( (3, 0) \) and \( (5, 0) \)
- y-intercept: \( (0, -15) \)
- vertex: \( (4, 1) \)
- Plot the points: Draw the x-intercepts, y-intercept, and vertex on a graph.
- Draw the parabola: Connect these points smoothly to form a U-shaped curve (in this case, an upside-down U due to the negative \( a \)). Ensure the vertex is the highest point, and the ends extend both leftwards and rightwards infinitely.
- Check your work: Make sure all calculated points and the direction of the parabola match the original quadratic function.