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Chapter 3: Problem 10

Graphing Quadratic Functions A quadratic function \(f\) is given. (a) Express \(f\) in standard form. (b) Find the vertex and \(x\) and \(y\) -intercepts of \(f .\) (c) Sketch a graph of \(f .\) (d) Find the domain and range of \(f\). $$f(x)=x^{2}+4 x-1$$

Short Answer

Expert verified
The vertex is \((-2, -5)\), x-intercepts are \(x = -2 \pm \sqrt{5}\), y-intercept is \((0, -1)\). Domain: \((- fty, fty)\), Range: \([-5, fty)\).

Step by step solution

01

Express in Standard Form

The given quadratic function is already in standard form, which is: \[ f(x) = ax^2 + bx + c \] For this function, \( a = 1 \), \( b = 4 \), and \( c = -1 \). Thus, the standard form of the function is: \[ f(x) = x^2 + 4x - 1 \]
02

Find the Vertex

To find the vertex of a quadratic function in standard form, use the formula for the vertex: \( x = \frac{-b}{2a} \). Substitute \( a = 1 \) and \( b = 4 \):\[ x = \frac{-4}{2 \times 1} = -2 \] To find the y-coordinate, substitute \( x = -2 \) back into the function:\[ f(-2) = (-2)^2 + 4(-2) - 1 = 4 - 8 - 1 = -5 \] Thus, the vertex is \( (-2, -5) \).
03

Find the x-intercepts

The x-intercepts are found by setting \( f(x) = 0 \) and solving for \( x \):\[ x^2 + 4x - 1 = 0 \] Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting \( a = 1 \), \( b = 4 \), \( c = -1 \):\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-1)}}{2 \times 1} = \frac{-4 \pm \sqrt{16 + 4}}{2} \]\[ x = \frac{-4 \pm \sqrt{20}}{2} = \frac{-4 \pm 2\sqrt{5}}{2} \]\[ x = -2 \pm \sqrt{5} \]Thus, the x-intercepts are \( x = -2 + \sqrt{5} \) and \( x = -2 - \sqrt{5} \).
04

Find the y-intercept

The y-intercept is found by setting \( x = 0 \) and solving for \( f(0) \):\[ f(0) = 0^2 + 4 \times 0 - 1 = -1 \] Thus, the y-intercept is \((0, -1)\).
05

Sketch the Graph

To sketch the graph of \( f(x) = x^2 + 4x - 1 \), plot the vertex \((-2, -5)\), the y-intercept \((0, -1)\), and the x-intercepts \((-2 + \sqrt{5}, 0)\) and \((-2 - \sqrt{5}, 0)\). Draw a parabola opening upwards, as the coefficient of \( x^2 \) is positive.
06

Determine the Domain and Range

The domain of any quadratic function is all real numbers, so the domain is \((-fty, fty)\). The range starts from the y-coordinate of the vertex for parabolas that open upwards. So, the range is \([-5, fty)\).

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Key Concepts

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

Standard Form
The standard form of a quadratic function is the universal way to express quadratic equations. It is written as:
  • \( f(x) = ax^2 + bx + c \)
where:
  • \( a \), \( b \), and \( c \) are constants.
  • \( a \) cannot be zero, because then it would not be a quadratic function.
In the exercise, we see the function \( f(x) = x^2 + 4x - 1 \). This is already in standard form, where \( a = 1 \), \( b = 4 \), and \( c = -1 \). This format is crucial for easily applying it to various operations, such as finding the vertex or intercepts.
Understanding the standard form helps simplify the process of analyzing and graphing quadratic functions. It provides an immediate view of the equation's structure.
Vertex of a Parabola
The vertex of a parabola is a fundamental point, representing its peak or trough. In mathematical terms, the vertex is the turning point where the parabola changes direction. For a function in standard form \( f(x) = ax^2 + bx + c \), you find the x-coordinate of the vertex using:
  • \( x = \frac{-b}{2a} \)
Once you have the x-coordinate, find the y-coordinate by substituting \( x \) back into the function. Let's see this in action for our function \( f(x) = x^2 + 4x - 1 \):
  • The vertex's x-coordinate is \( x = \frac{-4}{2 \times 1} = -2 \).
  • Substitute \( x = -2 \) into the function to find \( f(-2) = (-2)^2 + 4(-2) - 1 = -5 \).
Therefore, the vertex is at \((-2, -5)\). This point is crucial when graphing because it helps establish the basic shape and position of the parabola.
Graphing Functions
Graphing a quadratic function requires plotting key points and understanding their arrangement. Start with the vertex, as it dictates the parabola's direction. For our function, the vertex is \((-2, -5)\). You also need:
  • The y-intercept, where \( x = 0 \), found easily as \( f(0) = -1 \). So, the y-intercept is \((0, -1)\).
  • The x-intercepts, where \( f(x) = 0 \). We use the quadratic formula to determine that they are \( x = -2 \pm \sqrt{5} \).
With these points plotted, sketch a parabola starting from the vertex and passing through the intercepts. Make sure it opens upwards, as the coefficient of \( x^2 \) is positive in our example, ensuring the 'U' shape is correct.
Domain and Range
Understanding the domain and range of a quadratic function is key to fully grasping its behavior. The domain is quite straightforward: quadratic functions are defined for all real x-values. Hence, the domain is:
  • \((-\infty, \infty)\)
The range, however, depends on the parabola's direction. Since our function \( f(x) = x^2 + 4x - 1 \) opens upwards, the range starts from the vertex's y-coordinate and extends to infinity. Thus, the range is:
  • \([-5, \infty)\)
This tells us that the function's output values (y-values) begin at \(-5\) and increase indefinitely. Understanding these aspects reinforces our comprehension of the function's overall shape and scope.

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Most popular questions from this chapter

Give an example of a rational function that has vertical asymptote \(x=3 .\) Now give an example of one that has vertical asymptote \(x=3\) and horizontal asymptote \(y=2 .\) Now give an example of a rational function with vertical asymptotes \(x=1\) and \(x=-1,\) horizontal asymptote \(y=0,\) and \(x\) -intercept 4

Graphing Quadratic Functions A quadratic function \(f\) is given. (a) Express \(f\) in standard form. (b) Find the vertex and \(x\) and \(y\) -intercepts of \(f .\) (c) Sketch a graph of \(f .\) (d) Find the domain and range of \(f\). $$f(x)=-4 x^{2}-12 x+1$$

Graph the rational function, and find all vertical asymptotes, \(x\) - and \(y\) -intercepts, and local extrema, correct to the nearest tenth. Then use long division to find a polynomial that has the same end behavior as the rational function, and graph both functions in a sufficiently large viewing rectangle to verify that the end behaviors of the polynomial and the rational function are the same. $$y=\frac{x^{5}}{x^{3}-1}$$

Graphing Quadratic Functions A quadratic function \(f\) is given. (a) Express \(f\) in standard form. (b) Find the vertex and \(x\) and \(y\) -intercepts of \(f .\) (c) Sketch a graph of \(f .\) (d) Find the domain and range of \(f\). $$f(x)=2 x^{2}+12 x+10$$

Find the maximum or minimum value of the function. $$h(x)=\frac{1}{2} x^{2}+2 x-6$$
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