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Chapter 3: Problem 10

Graph the functions in Problems 10 through 18 by starting with the graph of a familiar function and applying appropriate shifts, flips, and stretches. Label all \(x\) - and \(y\) -intercepts and the coordinates of any vertices and corners. Use exact values, not numerical approximations. (a) \(y=(x-1)^{2}\) (b) \(y=-x^{2}-1\)

Short Answer

Expert verified
(a) Vertex: (1,0), Y-intercept: (0,1), X-intercept: (1,0). (b) Vertex: (0,-1), Y-intercept: (0,-1), no X-intercept

Step by step solution

01

Identifying the basic form of the function

The given function in (a) \(y=(x-1)^{2}\) is a quadratic function in standard form \(y=a(x-h)^{2}+k\). Here, \(a=1\), \(h=1\) and \(k=0\). In function (b) \(y=-x^{2}-1\), the \(a=-1\), \(h=0\) and \(k=-1\).
02

Determine the Vertex

The vertex of function (a) will be at the point (h, k) which is (1, 0). For function (b), the vertex will be at point (0,-1).
03

Calculate the y-intercepts

The y-intercepts can be found by setting x = 0. So for function (a), when \(x = 0, y = (0 - 1)^{2} = 1\). Therefore, the y-intercept is at point (0,1). For function (b), when \(x = 0, y = -0^{2}-1 = -1\). Therefore, the y-intercept is at point (0,-1).
04

Calculate the x-intercepts

The x-intercepts can be found by setting y = 0. For function (a), when \(y = 0, x = 1\). Therefore, the x-intercept is at point (1,0). For function (b), there are no x-intercepts because y does not equal zero for any value of x
05

Graphing the functions

The function (a) is graphed by drawing a parabolic curve opening upwards with the vertex at point (1,0), y-intercept at (0,1) and the x-intercept at (1,0). Function (b) is graphed by drawing a parabolic curve opening downwards with vertex at point (0,-1) and y-intercept at (0,-1). There is no x-intercept in function (b).

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Key Concepts

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

Function Transformation
Understanding how a function is transformed helps you easily graph it. Transformations can include shifts, stretches, compressions, and reflections.
For quadratic functions, these transformations change the shape and position of the parabola:
  • Shifts: The graph can move horizontally or vertically. For example, in function (a) \(y = (x-1)^2\), subtracting 1 from \(x\) shifts the parabola to the right by 1 unit.
  • Reflections: A negative sign in front of the squared term, as in function (b) \(y = -x^2 - 1\), flips the parabola vertically.
  • Stretches and Compressions: If \(a\) is greater than 1, the parabola becomes narrower. If \(a\) is between 0 and 1, it becomes wider.
These transformations allow you to predict and understand the behavior of the graph effectively.
Graphing Parabolas
Graphing a quadratic function involves sketching its parabolic shape. This is done by plotting key features like vertices, intercepts, and direction of opening.
To graph functions like those given, follow these steps:
  • Identify the vertex. It's a critical point where the parabola either peaks or troughs. In function (a), the vertex is \((1,0)\). In function (b), it's \((0,-1)\).
  • Determine the direction: If \(a > 0\), the parabola opens upwards. If \(a < 0\), it opens downwards, like in function (b).
  • Find intercepts: These points provide additional guidance for sketching. The \(y\)-intercepts are (0,1) and (0,-1) for functions (a) and (b) respectively. Function (a) has an \(x\)-intercept at (1,0), while function (b) has none.
Once outlined, connect the dots to form the smooth, U-shaped curve characteristic of a parabola.
Vertex of a Parabola
The vertex is the highest or lowest point on a parabola, depending on its direction. It provides a pivotal reference point for graphing.
For quadratic functions in the form \(y = a(x-h)^2 + k\), the vertex is at \((h, k)\).
  • In function (a) \(y = (x-1)^2\), the vertex is \((1, 0)\). This is found by identifying \(h = 1\) and \(k = 0\).
  • In function (b) \(y = -x^2 - 1\), the vertex is \((0, -1)\), where \(h = 0\) and \(k = -1\).
The position of the vertex helps determine the symmetry of the parabola and how transformations affect its graph. Knowing the vertex allows for accurate sketches and a better understanding of the function's behavior.

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Most popular questions from this chapter

Let \(j(x)=10\left(x^{-2}+2 x^{2}\right)^{3}\). Give two possible decompositions of \(j(x)\) such that \(j(x)=f(g(h(x)))\). None of the functions \(f, g\), and \(h\) should be the identity function.

Two brothers, Max and Eli, are experimenting with their walkie-talkies. (A walkie-talkie is a combined radio transmitter and receiver light enough to allow the user to walk and talk at the same time.) The quality of the transmission, \(Q\), is a function of the distance between the two walkie- talkies. We will model it as being inversely proportional to this distance. At time \(t=0\) Max is 100 feet north of Eli. Max walks north at a speed of 300 feet per minute while Eli walks east at a speed of 250 feet per minute. All the time they are talking on their walkie-talkies. (a) Write a function \(f\) such that \(Q=f(d)\), where \(d\) is the distance between the brothers. Your function will involve an unknown constant. (b) Write a function \(g\) that gives the distance between the brothers at time \(t\). (c) Find \(f(g(t))\). What does this composite function take as input and what does it give as output?

Find functions \(f\) and \(g\) such that \(h(x)=f(g(x))\) and neither \(f\) nor \(g\) is the identity function, i.e., \(f(x) \neq x\) and \(g(x) \neq x .\) Answers to these problems are not unique. \(h(x)=5(x-\pi)^{2}+4(x-\pi)+7\)

For each of the functions given below, give possible formulas for \(f(x)\) and \(g(x)\) such that \(h(x)=f(g(x))\). Do not let \(g(x)=x\); do not let \(f(x)=x\). (a) \(h(x)=\sqrt{x^{2}+3}\) (b) \(h(x)=\sqrt{x}+\frac{5}{\sqrt{x}}\) (c) \(h(x)=\frac{3}{3 x^{2}+2 x}\) (d) \(h(x)=5\left(x^{2}+3 x^{3}\right)^{3}\)

Find functions \(f\) and \(g\) such that \(h(x)=f(g(x))\) and neither \(f\) nor \(g\) is the identity function, i.e., \(f(x) \neq x\) and \(g(x) \neq x .\) Answers to these problems are not unique. \(h(x)=\frac{x+1}{x+2}\)
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