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Chapter 4: Problem 52

(a) find the vertex and the axis of symmetry of each quadratic function, and determine whether the graph is concave up or concave down. (b) Find the y-intercept and the \(x\) -intercepts, if any. (c) Use parts (a) and (b) to graph the function. (d) Find the domain and the range of the quadratic function. (e) Determine where the quadratic function is increasing and where it is decreasing. (f) Determine where \(f(x)>0\) and where \(f(x)<0\) \(f(x)=4 x^{2}-2 x+1\)

Short Answer

Expert verified
The vertex is \( (\frac{1}{4}, \frac{3}{4}) \), axis of symmetry \( x = \frac{1}{4} \); the graph is concave up with y-intercept at \( (0, 1) \), no x-intercepts, and the domain \( (-\infty, \infty) \) and range \( [\frac{3}{4}, \infty) \). The function decreases on \( (-\infty, \frac{1}{4}) \) and increases on \( (\frac{1}{4}, \infty) \). \( f(x) > 0 \) always.

Step by step solution

01

Finding the Vertex

The vertex of a quadratic function in the form \( f(x) = ax^2 + bx + c \) can be found using the formula for the x-coordinate: \( x_v = -\frac{b}{2a} \). In this problem, \( a = 4 \) and \( b = -2 \). Thus,\( x_v = -\frac{-2}{2 \cdot 4} = \frac{2}{8} = \frac{1}{4} \). Now, substitute \( x_v = \frac{1}{4} \) back into the function to find the y-coordinate of the vertex:\( f(\frac{1}{4}) = 4(\frac{1}{4})^2 - 2(\frac{1}{4}) + 1 = 4(\frac{1}{16}) - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}\). Therefore, the vertex is \((\frac{1}{4}, \frac{3}{4})\).
02

Axis of Symmetry

The axis of symmetry for a quadratic function is given by the equation \( x = x_v \). From Step 1, \( x_v = \frac{1}{4} \). Hence, the axis of symmetry is \( x = \frac{1}{4} \).
03

Concavity (Concave Up or Down)

The concavity of a quadratic function is determined by the sign of the coefficient \( a \). If \( a > 0 \), the graph is concave up. If \( a < 0 \), the graph is concave down. In \( f(x) = 4x^2 - 2x + 1 \), \( a = 4 \) which is greater than 0, so the graph is concave up.
04

Finding the y-intercept

The y-intercept of a quadratic function is found by setting \( x = 0 \) and solving for \( f(x) \). Here, \( f(0) = 4(0)^2 - 2(0) + 1 = 1 \). Therefore, the y-intercept is \( (0, 1) \).
05

Finding the x-intercepts

The x-intercepts are found by setting \( f(x) = 0 \) and solving for \( x \):\( 4x^2 - 2x + 1 = 0 \). This equation does not have real solutions since its discriminant \( \Delta = b^2 - 4ac = (-2)^2 - 4(4)(1) = 4 - 16 = -12 \) is negative. Hence, there are no x-intercepts.
06

Graphing the Function

Using the information from Steps 1 to 5: - Vertex: \( (\frac{1}{4}, \frac{3}{4}) \)- Axis of symmetry: \( x = \frac{1}{4} \)- y-intercept: \( (0, 1) \)- No x-intercepts- Concave up Graph these points and sketch the parabola opening upwards.
07

Domain of the Function

The domain of any quadratic function is all real numbers. Therefore, the domain is \( (-\infty, \infty) \).
08

Range of the Function

Since the quadratic function is concave up and has a vertex at \( (\frac{1}{4}, \frac{3}{4}) \), the minimum y-value is \( \frac{3}{4} \). Thus, the range is \( [\frac{3}{4}, \infty) \).
09

Intervals of Increase and Decrease

The function decreases on the interval \( (-\infty, \frac{1}{4}) \) and increases on the interval \( (\frac{1}{4}, \infty) \).
10

Where \( f(x) > 0 \) and \( f(x) < 0 \)

Since the parabola opens upwards and the minimum value of \( f(x) \) is \( \frac{3}{4} \), \( f(x) > 0 \) for all \( x \). Therefore, \( f(x) < 0 \) for no values of \( x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

vertex of a quadratic function
The vertex of a quadratic function gives you the highest or lowest point on a parabola. For a function in standard form, \( f(x) = ax^2 + bx + c \), you can find the x-coordinate of the vertex using the formula: \( x_v = -\frac{b}{2a} \).
After you've found the x-coordinate, substitute it back into the function to find the y-coordinate, giving you the vertex \( (x_v, y_v) \). For our example, the vertex is located at \( \left(\frac{1}{4}, \frac{3}{4}\right) \).
This is a crucial point as it helps determine many other properties of the function.
axis of symmetry
The axis of symmetry is a vertical line that divides the parabola into mirror images. It runs through the vertex and has the equation \( x = x_v \).
From our earlier calculations, the axis of symmetry for our function is \( x = \frac{1}{4} \).
This line helps to verify the shape and positioning of the parabola.
parabola concavity
Concavity refers to whether the parabola opens upwards or downwards. It's determined by the coefficient \( a \) in the quadratic function.
If \( a > 0 \), the parabola is concave up, resembling a 'U' shape. If \( a < 0 \), it's concave down, resembling an upside-down 'U'.
Since our function has \( a = 4 \) (which is greater than 0), it opens upward, making it concave up.
y-intercept
The y-intercept is the point where the graph crosses the y-axis, found by setting \( x = 0 \) and solving for \( f(x) \).
In our function, \( f(0) = 1 \), so the y-intercept is \( (0, 1) \).
This point helps to initialize the graph, especially when plotting by hand.
x-intercepts
The x-intercepts are where the graph crosses the x-axis, found by setting \( f(x) = 0 \) and solving for \( x \).
In our example, the equation \( 4x^2 - 2x + 1 = 0 \) has no real solutions since its discriminant is negative (\( \triangle = -12 \)).
This means there are no x-intercepts, and the parabola does not touch the x-axis.
domain and range
The domain of any quadratic function is all real numbers: \( (-\infty, \infty) \).
The range depends on the vertex and the direction the parabola opens (concavity). Since our parabola opens upwards and the vertex is at \( \left(\frac{1}{4}, \frac{3}{4}\right) \), the minimum y-value is \( \frac{3}{4} \).
Therefore, the range is \( \left[\frac{3}{4}, \infty\right) \).
increasing and decreasing intervals
For quadratic functions, the graph is split into increasing and decreasing intervals by the vertex.
To the left of the vertex, the function decreases, and to the right, it increases. Using our function's vertex at \( \left(\frac{1}{4}, \frac{3}{4}\right) \), it decreases on \( (-\infty, \frac{1}{4}) \) and increases on \( (\frac{1}{4}, \infty) \).
These intervals help understand the behavior of the function over different x-values.

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