Chapter 2: Problem 18
(a) Use the quadratic formula to find the zeros of \(f .\) (b) Find the maximum or minimum value of \(f(x)\) (c) Sketch the graph of \(f\) $$f(x)=-4 x^{2}+4 x-1$$
Short Answer
Expert verified
The zero is \( x = \frac{1}{2} \), the maximum value is 0, and the graph is a downward-opening parabola peaking at the vertex \( (\frac{1}{2}, 0) \).
Step by step solution
01
Determine the Coefficients
To use the quadratic formula, identify the coefficients of the quadratic equation \( f(x) = -4x^2 + 4x - 1 \). Here, \( a = -4 \), \( b = 4 \), and \( c = -1 \).
02
Apply the Quadratic Formula
Substitute the coefficients into the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Plug in the values: \[ x = \frac{-4 \pm \sqrt{4^2 - 4(-4)(-1)}}{2(-4)} \]This simplifies to: \[ x = \frac{-4 \pm \sqrt{16 - 16}}{-8} \]Thus, \( x = \frac{-4 \pm \sqrt{0}}{-8} \), which further simplifies to \( x = \frac{-4}{-8} \) or \( x = \frac{1}{2} \).
03
Find the Vertex
The vertex form of a quadratic function is given by \( f(x) = ax^2 + bx + c = a(x-h)^2 + k \), where \( (h, k) \) is the vertex and the value at \( h \) gives the maximum or minimum. Calculate \( h \) using \( h = -\frac{b}{2a} \):\[ h = -\frac{4}{2(-4)} = \frac{1}{2} \]Now, find \( k \) by substituting \( h \) back into \( f(x) \):\[ k = f\left(\frac{1}{2}\right) = -4\left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right) - 1 \]\[ k = -4 \cdot \frac{1}{4} + 2 - 1 = -1 + 2 - 1 = 0 \]Therefore, the vertex is \( (\frac{1}{2}, 0) \) and it provides a maximum value of 0.
04
Sketch the Graph
Using the vertex \( (\frac{1}{2}, 0) \) and the information that the graph opens downwards (since \( a = -4 \), which is negative), plot the graph. The zero found earlier at \( x = \frac{1}{2} \) makes the vertex the x-intercept. Draw the parabola showing it opening downwards and peaking at the vertex.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \( f(x) = ax^2 + bx + c \). If you want to find the zeros of a quadratic function, meaning the points where its graph crosses the x-axis, the quadratic formula can help. It's expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula helps you determine the values of \( x \) when \( f(x) = 0 \). Here's how you can apply it:
- Identify the coefficients \( a \), \( b \), \( c \) from your quadratic function. For example, in \( f(x) = -4x^2 + 4x - 1 \), we have \( a = -4 \), \( b = 4 \), and \( c = -1 \).
- Plug these values into the formula to find the roots, \( x = \frac{-4 \pm \sqrt{16 - 16}}{-8} \), which simplifies to \( x = \frac{1}{2} \). In this case, the discriminant \( \sqrt{b^2 - 4ac} \) is zero, indicating one repeated root.
Vertex of a Parabola
Finding the vertex of a parabola is crucial, especially when determining whether the parabola has a maximum or minimum value. The vertex can be understood as the peak (maximum) or the trough (minimum) of the graph. For a parabola given as \( ax^2 + bx + c \), the x-coordinate of the vertex \( (h,k) \) can be found using:\[ h = -\frac{b}{2a}\]After calculating \( h \), substitute it back into the function to find \( k \).
- In \( f(x) = -4x^2 + 4x - 1 \), calculate \( h \): \( h = -\frac{4}{2(-4)} = \frac{1}{2} \).
- Use this value to find \( k \): \( k = f\left(\frac{1}{2}\right) = 0 \), making the vertex \( \left(\frac{1}{2}, 0\right) \).
Graphing Quadratics
Graphing quadratic functions involves plotting a parabola on a coordinate plane. These functions can be graphed systematically by understanding their key features: the vertex, axis of symmetry, direction, and intercepts.To start, identify the vertex, which is \( \left(\frac{1}{2}, 0\right) \) for the function \( f(x) = -4x^2 + 4x - 1 \). This tells us the highest point on the graph since the parabola opens downward.Several steps are involved in sketching the graph:
- Vertex: Plot the vertex point on the graph.
- Axis of Symmetry: Draw a vertical line through the vertex. The parabola is symmetric about this line.
- Direction: Since \( a = -4 \) is negative, the parabola opens downward.
- Zero: As solved using the quadratic formula, the parabola touches the x-axis at \( x = \frac{1}{2} \). This intersection is both a vertex and a root.