Explanation Given information: A function is given as, f ( x ) = x 2 + x − 1 . Three lines are given as, y = x + 3 y = x − 1 and y = x − 3 Calculation: Three lines are given as, y = x + 3 ... eq(i) y = x − 1 ... eq(ii) and y = x − 3 ... eq(iii) Given function is, f ( x ) = x 2 + x − 1 . Let us suppose, y = f ( x ) = x 2 + x − 1 ...eq(iv) If the line y = x + 3 passes through the graph of function f ( x ) , then upon solving eq(i) and eq(iv) will give the point of intersection. Substituting y = x + 3 in eq(iv), y = x 2 + x − 1 ⇒ x + 3 = x 2 + x − 1 ⇒ x 2 = 4 ⇒ x = ± 2 From eq(i) y = x + 3 , when x = 2 , then y = 2 + 3 = 5 when x = − 2 , then y = − 2 + 3 = 1 Therefore, point of intersection will be ( 2 , 5 ) and ( − 2 , 1 ) Again, if the line y = x − 3 passes through the graph of function f ( x ) , then upon solving eq(ii) and eq(iv) will give the point of intersection To determine (b) The sketch of function f and three given lines. To determine (c) The relationship between graph of f and the lines y = x + c as c changes from − ∞ to ∞ .

4th Edition

ISBN: 9781319545956

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 1.2, Problem 50E

To determine

(a)

To show:

The lines y=x+3,y=x−1, and y=x−3 intersect the graph of f in two, one, and zero points respectively.

To determine

(b)

The sketch of function f and three given lines.

To determine

(c)

The relationship between graph of f and the lines y=x+c as c changes from −∞ to ∞.

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Chapter 1.2, Problem 49E

Chapter 1.2, Problem 51E

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A factorization A = PDP 1 is not unique. For A= 7 2 -4 1 1 1 5 0 2 1 one factorization is P = D= and P-1 30 = Use this information with D₁ = to find a matrix P₁ such that - -1 -2 0 3 1 - - 1 05 A-P,D,P P1 (Type an integer or simplified fraction for each matrix element.)

Matrix A is factored in the form PDP 1. Use the Diagonalization Theorem to find the eigenvalues of A and a basis for each eigenspace. 30 -1 - 1 0 -1 400 0 0 1 A= 3 4 3 0 1 3 040 3 1 3 0 0 4 1 0 0 003 -1 0 -1 Select the correct choice below and fill in the answer boxes to complete your choice. (Use a comma to separate vectors as needed.) A basis for the corresponding eigenspace is { A. There is one distinct eigenvalue, λ = B. In ascending order, the two distinct eigenvalues are λ₁ ... = and 2 = Bases for the corresponding eigenspaces are { and ( ), respectively. C. In ascending order, the three distinct eigenvalues are λ₁ = = 12/2 = and 3 = Bases for the corresponding eigenspaces are {}, }, and { respectively.

Chapter 1 Solutions

CALCULUS W/ACHIEVE PLATFORM

Ch. 1.1 - Prob. 1PQ Ch. 1.1 - Prob. 2PQ Ch. 1.1 - Prob. 3PQ Ch. 1.1 - Prob. 4PQ Ch. 1.1 - Prob. 5PQ Ch. 1.1 - Prob. 6PQ Ch. 1.1 - Prob. 7PQ Ch. 1.1 - Prob. 8PQ Ch. 1.1 - Prob. 9PQ Ch. 1.1 - Prob. 10PQ

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(a) To show: The lines y = x + 3 , y = x − 1 , and y = x − 3 intersect the graph of f in two, one, and zero points respectively.

Solution Summary: The author explains the function f(x)=x2+x-1.

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(a)

To show:

The lines y=x+3,y=x−1, and y=x−3 intersect the graph of f in two, one, and zero points respectively.

(b)

The sketch of function f and three given lines.

(c)

The relationship between graph of f and the lines y=x+c as c changes from −∞ to ∞.

Want to see the full answer?

Chapter 1 Solutions

(a) To show: The lines y = x + 3 , y = x − 1 , and y = x − 3 intersect the graph of f in two, one, and zero points respectively.

Solution Summary: The author explains the function f(x)=x2+x-1.

Concept explainers

Videos

(a) To show: The lines y=x+3,y=x−1, and y=x−3 intersect the graph of f in two, one, and zero points respectively.

(b) The sketch of function f and three given lines.

(c) The relationship between graph of f and the lines y=x+c as c changes from −∞ to ∞.

Want to see the full answer?

Chapter 1 Solutions

(a)

To show:

The lines y=x+3,y=x−1, and y=x−3 intersect the graph of f in two, one, and zero points respectively.

(b)

The sketch of function f and three given lines.

(c)

The relationship between graph of f and the lines y=x+c as c changes from −∞ to ∞.