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Chapter 4: Problem 55

(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 \(f(x)<0\) \(f(x)=3 x^{2}+6 x+2\)

Short Answer

Expert verified
Vertex: (-1, -1). Axis: x = -1. Concave up. y-intercept: (0, 2). x-intercepts: -1 +/- (sqrt(3))/3. Domain: (-infinity, infinity). Range: [-1, infinity). Increasing: (-1, infinity). Decreasing: (-infinity, -1).

Step by step solution

01

Find the Vertex

To find the vertex of the quadratic function, first use the formula for the x-coordinate of the vertex, which is \[ x = -\frac{b}{2a} \]For the given function, \( a = 3 \) and \( b = 6 \).Substituting these values in, we get:\[ x = -\frac{6}{2 \times 3} = -1 \]Next, substitute \( x = -1 \) back into the equation to find the y-coordinate of the vertex: \[ f(-1) = 3(-1)^{2} + 6(-1) + 2 = 3 - 6 + 2 = -1 \]Thus, the vertex is \[ (-1, -1) \].
02

Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex. It can be written as:\[ x = -1 \].
03

Determine Concavity

The concavity of the graph is determined by the coefficient of the \( x^2 \) term, which is \( a \). If \( a > 0 \), the parabola opens upwards (concave up). If \( a < 0 \), the parabola opens downwards (concave down). Here, \( a = 3 > 0 \), so the graph is concave up.
04

Find the y-intercept

The y-intercept is found by setting \( x = 0 \). Substituting \( x = 0 \) into the function gives:\[ f(0) = 3(0)^2 + 6(0) + 2 = 2 \]So, the y-intercept is \( (0, 2) \).
05

Find the x-intercepts

The x-intercepts are found by setting \( f(x) = 0 \) and solving for \( x \):\[ 3x^2 + 6x + 2 = 0 \]Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 6 \), and \( c = 2 \), we get:\[ x = \frac{-6 \pm \sqrt{36 - 24}}{6} = \frac{-6 \pm 2\sqrt{3}}{6} = \frac{-3 \pm \sqrt{3}}{3} \]Thus, the x-intercepts are \( x = -1 + \frac{\sqrt{3}}{3} \) and \( x = -1 - \frac{\sqrt{3}}{3} \).
06

Graph the Function

Using the information from steps 1-5, plot the vertex \((-1, -1)\), axis of symmetry \( x = -1 \), y-intercept \((0, 2)\), and x-intercepts \( x = -1 + \frac{\sqrt{3}}{3} \) and \( x = -1 - \frac{\sqrt{3}}{3} \). Draw the parabola opening upwards.
07

Find the Domain and Range

The domain of any quadratic function is all real numbers, so:\[ \text{Domain} = (-\infty, \infty) \]The range depends on the vertex and the direction of concavity. Since the parabola opens upwards and the vertex is \((-1, -1)\), the range is:\[ \text{Range} = [-1, \infty) \]
08

Determine Increasing and Decreasing Intervals

The function is increasing when the values of \( f(x) \) get larger as \( x \) increases, and decreasing when \( f(x) \) values become smaller as \( x \) increases. Since the vertex is the lowest point:- The function is decreasing on the interval \( (-\infty, -1) \)- The function is increasing on the interval \( (-1, \infty) \)
09

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

The function is greater than zero between the x-intercepts and less than zero outside the x-intercepts:- \( f(x) > 0 \) on the intervals \( \left( -1 - \frac{\sqrt{3}}{3}, -1 + \frac{\sqrt{3}}{3} \right) \)- \( f(x) < 0 \) on the intervals \( (-\infty, -1 - \frac{\sqrt{3}}{3}) \) and \( (-1 + \frac{\sqrt{3}}{3}, \infty) \)

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Key Concepts

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

vertex
The vertex of a quadratic function is a pivotal point that defines the function鈥檚 peak or trough. This is found using the formula for the x-coordinate: \( x = -\frac{b}{2a} \). For our function \( f(x) = 3x^2 + 6x + 2 \), we get \( x = -1 \).To find the y-coordinate, substitute \( x = -1 \) back into the function: \( f(-1) = 3(-1)^2 + 6(-1) + 2 = -1 \). Thus, the vertex is at \( (-1, -1) \).
axis of symmetry
The axis of symmetry for a quadratic function is a vertical line that passes through the vertex, splitting the parabola into two mirrored halves. For the function \( f(x) = 3x^2 + 6x + 2 \), we found the vertex to be at \( (-1, -1) \), so the axis of symmetry is: \( x = -1 \).
concavity
Concavity tells us whether the parabola opens upwards or downwards. This is determined by the coefficient of the \( x^2 \) term, which is \( a \). If \( a > 0 \), the parabola is concave up. If \( a < 0 \), it is concave down. In our function, \( a = 3 \), which is greater than zero, so the graph is concave up.
y-intercept
The y-intercept is where the graph crosses the y-axis. This is found by setting \( x = 0 \). For the function \( f(x) = 3x^2 + 6x + 2 \), substituting \( x = 0 \) gives us: \( f(0) = 2 \). So, the y-intercept is at \( (0, 2) \).
x-intercepts
The x-intercepts, also known as the roots or zeros of the function, are the points where the graph crosses the x-axis. These are found by setting \( f(x) = 0 \) and solving for \( x \). For \( f(x) = 3x^2 + 6x + 2 \), use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting \( a = 3 \), \( b = 6 \) and \( c = 2 \), we get: \( x = -1 \pm \frac{\sqrt{3}}{3} \). Thus, the x-intercepts are: \( x = -1 + \frac{\sqrt{3}}{3} \) and \( x = -1 - \frac{\sqrt{3}}{3} \).
domain and range
The domain of any quadratic function is all real numbers, \( (-\infty, \infty) \). The range is based on the vertex and direction of the parabola. Our function opens upwards and has a vertex at \( (-1, -1) \), so the range is: \( [-1, \infty) \).
increasing and decreasing intervals
For quadratic functions, the graph increases and decreases around the vertex. Since our vertex is \( (-1, -1) \) and the parabola opens upwards:
  • The function decreases on \( (-\infty, -1) \)
  • It increases on \( (-1, \infty) \).
Understanding these intervals helps in sketching the graph and analyzing the function's behavior.

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