Question

6.

2.Find the triangle bounded by the y-axis the line f(x)=9- 6/7x and the line perpendicular to it that passes through the origin

Answer 1

Hello,

Please, see the attached files.

Thanks.

Answer 2

First.

We need to find the triangle bounded by the y-axis the line
`f(x)=9-(6)/(7)x` and the line perpendicular to it that passes through the origin.

This triangle is illustrated in the First Figure bellow. The triangle has been dotted with black dots. So, let's identify each part of this graph:

• The y-axis has been identified by the green line.
• The line
  `f(x)=9-(6)/(7)x` has been identified by the red line.
• To plot the line perpendicular to the previous one that passes through the origin, we need to know that the product of the slopes of two perpendicular lines is
  `m_(1)m_(2)=-1 \therefore Since \ m_(1)=-(6)/(7) \ then \ m_(2)=(7)/(6)`

Therefore, this line is the blue one and is written as:

`f(x)=(7)/(6)x`

Second.

a.

To find
`y(t)` we need to find the slope and y-intercept. So, taking two points:

`P_(1)(15,150) \ and \ P_(2)(25,450) \\ \\m=(450-150)/(25-15)=30`

So:

`Using \ P_(1)(15,150) \\\\y=30t+b\\\\150=30(15)+b \therefore b=-300 \\\\\boxed{y(t)=30t-300}`

Third.

b.

The y-intercept is
`b=-300` and has been calculated above. This value represents the profit in thousands of dollars in 1980, that is, there is no any profit or there is a lost of $300.000.

Fourth.

c.

The x-intercept can be found when y=0 (year 1980), that is:

`y(t)=30t-300 \\\\0=30t-300 \therefore t=10`

This value shows that there is no any profit for the company at the year that represents t=10 (year 1990), because at this point the company starts earning money.

Fifth.

d.

The slope of this function is
`m=30`

This value tells you that the profit increases $30.000 each year.