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Graphing equations: Graph the following equations. Use a

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 7RE Chapter 1

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 7RE

Graphing equations: Graph the following equations. Use a graphing utility only to check your work. ? a?. 2?x ? 3y ? + 10 = 0 ? ? b.??y = ?x2+ 2?x ? 3 ? ? c? . ?x2 +? 2x? + y ? 2 + 4y ? + 1 = 0 ? ? ? d? .? 7??? 2x ? + y?2 ? 8?y + 5 = 0

Step-by-Step Solution:

Step-by-step solution Step 1 a) We need to graph the given function 2x3y+10 = 0. Step 2 From the degree of the variables, we know that the given function is linear. With this, we express the function in Slope-Intercept Form. 3y = 2x+10 2 10 y = 3+ 3 Step 3 Next, to make the graphing easier, we find the x and y-intercepts of the function. At x = 0 : y = (0)+ 10 3 3 y = 10 3 10 yintercept is (0, 3) At y = 0 : 2 10 0 = 3+ 3 2x =10 x =5 xintercept is (5, 0) Step 4 Using the intercepts and graphing, we get: Step 5 b) We need to graph the given function y = x +2x3. Step 6 From the degree of the variables, we know that the given function is quadratic and the graph is a parabola. With this, we find the vertex and the intercepts of the function to make the graphing easier. To find the vertex, we express the given function in V ertex Form. h = 2a = 2 =1 2 k = f(h) = (1) +2(1)3 =4 y = (x+1) 4 Therefore: vertex is at (1, 4).

Step 7 of 12

Chapter 1, Problem 7RE is Solved
Step 8 of 12

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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