   Chapter 5.5, Problem 17E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding the Area Bounded by Two Graphs In Exercises 15-30, sketch the region bounded by the graphs of the functions and find the area of the region. See Examples 1, 2, 3, and 4. y = x 2 − 4 x + 3 , y = 3 + 4 x − x 2

To determine

To graph: The region bounded by the graphs of y=x24x+3 and y=3+4xx2, and also compute the area of the region.

Explanation

Given Information:

The region bounded by the graphs of y=x24x+3 and y=3+4xx2.

Graph:

Consider the provided functions,

y=x24x+3 and y=3+4xx2

Then the first function y=x24x+3 is a quadratic function. So, its graph will be a parabola with opening upwards.

Factor the function,

y=x24x+3=x23xx+3=x(x3)1(x3)=(x3)(x1)

The x-intercepts are x=3 and x=1.

Consider the second function y=3+4xx2

It is a quadratic function. The term x2 is negative, so its graph will be a parabola with opening downwards. Compute its x-intercepts as follows,

3+4xx2=0x=4±424×3×(1)2×(1)=4±28(2)=2±7

The x-intercepts are x=2±7.

Compute the points of intersection of two graphs by setting the functions equal to each other and solving for x,

x24x+3=3+4xx2x2+x24x4x+33=02x28x=02x(x4)=0

Apply zero product property,

x=0x4=0x=4

Substitute x=0 in the function y=x24x+3 and compute the first intersection point,

y=024×0+3=00+3=3

Substitute x=4 in the function y=x24x+3 and compute the second intersection point,

y=424×4+3=1616+3=3

So, the graphs of given functions intersect at the points (0,3) and (4,3)

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