Explanation Given information: A function is given as, f ( x ) = x 2 + 2 x − 21 . Three lines are given as, y = 3 x − 25 y = 6 x − 25 and y = 9 x − 25 Calculation: Three lines are given as, y = 3 x − 25 ... eq(i) y = 6 x − 25 ... eq(ii) and y = 9 x − 25 ... eq(iii) Given function is, f ( x ) = x 2 + 2 x − 21 . Let us suppose, y = f ( x ) = x 2 + 2 x − 21 ...eq(iv) If the line y = 3 x − 25 passes through the graph of function f ( x ) , then upon solving eq(i) and eq(iv) will give the point of intersection. Substituting y = 3 x − 25 in eq(iv), y = x 2 + 2 x − 21 ⇒ 3 x − 25 = x 2 + 2 x − 21 ⇒ x 2 − x + 4 = 0 Here, b 2 − 4 a c = ( − 1 ) 2 − 4 ⋅ 1 ⋅ 4 = − 15 < 0 { From properties of quadratic equation } No real solution of above quadratic equation . Therefore, no point of intersection will be there . Again, if the line y = 6 x − 25 passes through the graph of function f ( x ) , then upon solving eq(ii) and eq(iv) will give the point of intersection. Substituting y = 6 x − 25 in eq(iv), y = x 2 + 2 x − 21 ⇒ 6 x − 25 = x 2 + 2 x − 21 ⇒ x 2 − 4 x + 4 = 0 ⇒ ( x − 2 ) 2 = 0 ⇒ x − 2 = 0 ⇒ x = 2 From eq(ii), y = 6 x − 25 ⇒ y = 6 ( 2 ) − 25 = − 13 Therefore, one point of intersection will be ( 2 , − 13 ) 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 = c x − 25 as c changes from 0 to ∞ .

4th Edition

ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Domain

In simple mathematics terms, a domain is defined as the space on which all the possible inputs for the function will lie. It can also be expressed as a set of departures for the required function.

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Question

Chapter 1.2, Problem 51E

To determine

(a)

To show:

The lines

y=3x−25,y=6x−25, and y=9x−25 intersect the graph of f in zero, one, and two 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=cx−25 as c changes from 0 to ∞.

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Chapter 1.2, Problem 50E

Chapter 1.2, Problem 52E

Chapter 1 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

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(a) To show: The lines y = 3 x − 25 , y = 6 x − 25 , and y = 9 x − 25 intersect the graph of f in zero, one, and two points respectively.

Solution Summary: The author illustrates how the lines y=3x-25,y =6x 25,andy>9x25 intersect the graph of in zero, one, and two points.

Concept explainers

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(a)

To show:

The lines

y=3x−25,y=6x−25, and y=9x−25 intersect the graph of f in zero, one, and two points respectively.

(b)

The sketch of function f and three given lines.

(c)

The relationship between graph of f and the lines y=cx−25 as c changes from 0 to ∞.

Want to see the full answer?

Chapter 1 Solutions

(a) To show: The lines y = 3 x − 25 , y = 6 x − 25 , and y = 9 x − 25 intersect the graph of f in zero, one, and two points respectively.

Solution Summary: The author illustrates how the lines y=3x-25,y =6x 25,andy>9x25 intersect the graph of in zero, one, and two points.

Concept explainers

Videos

(a) To show: The lines y=3x−25,y=6x−25, and y=9x−25 intersect the graph of f in zero, one, and two points respectively.

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

(c) The relationship between graph of f and the lines y=cx−25 as c changes from 0 to ∞.

Want to see the full answer?

Chapter 1 Solutions

(a)

To show:

The lines

y=3x−25,y=6x−25, and y=9x−25 intersect the graph of f in zero, one, and two points respectively.

(b)

The sketch of function f and three given lines.

(c)

The relationship between graph of f and the lines y=cx−25 as c changes from 0 to ∞.