Question

A right triangle is formed in the first quadrant by the x- and y-axes and a line through the point (5, 3), as shown in the figure below. Write the area A of the triangle as a function of x, and determine the domain of the function. (Enter your answer for the domain using interval notation.)

Answer 1

The area of the triangle as a function of x is A = 3x^2/[2(x -5)] and the domain is x > 5

Write the area A of the triangle as a function of x

From the figure, we have the following points:

(0,y), (5,3), and (x,0)

Next, we calculate the slopes between the points.

This is calculated as follows:

• Slope between (0,y) and (5,3) = [y - 3]/[0 - 5] = [3 - y]/5
• Slope between (5,3) and (x,0) = [3 - 0]/[5 - x] = 3/[5 - x]

The slopes are equal.

So, we have:

[3 - y]/5 = 3/[5 - x]

Cross multiply

(3 - y)(5 -x) = 15

Divide by 5 - x

(3 - y) = 15/(5 -x)

This gives

y = 3 - 15/(5 -x)

This gives

y = [15- 3x - 15]/[5 - x]

Evaluate

y = 3x/(x - 5)

The area is calculated as:

Area = 1/2 * xy

So, we have:

Area = 1/2 * x * 3x/(x - 5)

Evaluate

Area = 3x^2/[2(x -5)]

Hence, the area of the triangle as a function of x is A = 3x^2/[2(x -5)]

The domain of the function

We have:

A = 3x^2/[2(x -5)]

Set the denominator > 0

2(x - 5) > 0

Divide by 2

x - 5 > 0

Add 5 to the sides

x > 5

Hence, the domain is x > 5

Read more about domains at:

brainly.com/question/2428614

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