Question

Consider the equation of the line, c(c2 + 2)y = c − x where c > 0 is a constant. (a) Find the coordinates of the x-intercept and the y-intercept. x-intercept ( , ) y-intercept ( , ) (b) Find a formula for the area of the triangle enclosed between the line, the x-axis, and the y-axis. A(c) =

Answer 1

Answer:

Part a)

The y-intercept is the point $(0,\frac{1}{c^{2}+2})$

The x-intercept is the point $(c,0)$

Part b) The formula of the area of triangle is equal to

$A(c)=(\frac{c}{2c^{2}+4})$

Step-by-step explanation:

step 1

Find the coordinates of the x-intercept and the y-intercept

we have that

The equation of the line is equal to

$c(c^{2}+2)y=c-x$

Find the y-intercept (value of y when the value of x is equal to zero)

For x=0

$c(c^{2}+2)y=c-0$

$c(c^{2}+2)y=c$

simplify

$(c^{2}+2)y=1$

$y=\frac{1}{c^{2}+2}$

The y-intercept is the point $(0,\frac{1}{c^{2}+2})$

Find the x-intercept (value of x when the value of y is equal to zero)

For y=0

$c(c^{2}+2)(0)=c-x$

$0=c-x$

$x=c$

The x-intercept is the point $(c,0)$

step 2

Find the formula for the area of triangle enclosed between the line, the x-axis, and the y-axis

The formula for the area of triangle is equal to

$A=\frac{1}{2}(b)(h)$

In this problem

$b=c$ -----> the x-intercept

$h=\frac{1}{c^{2}+2}$ ----> the y-intercept

substitute

$A(c)=\frac{1}{2}(c)(\frac{1}{c^{2}+2})$

$A(c)=(\frac{c}{2c^{2}+4})$

Answer 2

The x-intercept and the y-intercept in the equation will be (c, 0) and (0, 1/c² + 2) respectively.

How to compute the equation?

The equation of the line is given as c(c² + 2)y = c - x.

The y-intercept will be:.

c(c² + 2)y = c - x

c(c² + 2)y = c - 0

c(c² + 2)y = c

y = 2/(c2 + 2)

The x-intercept will be:

c(c² + 2)y = c - x

0 = c - x

c = x

Therefore, the x-intercept is (c, 0).

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