Question

Can someone answer please?

Show full working out please

Answer 1

see explanation

Explanation:

a) vertical asymptote is where the slope of the function is vertical. This occurs where f(x) runs off to plus or minus infinity, which occurs when you divide by zero. This, in turn, occurs when x -> -3

So you can write as the equation of the vertical asymptote: `x = -3`

b) x axis intercepts occur where y = 0, so to find these, solve:

`(3x-2)/(x+3) = 0`

...and this happens when the numerator is zero, so :
`3x-2 = 0`
`3x = 2`
`x = 2/3`

So the x-axis intercept is `(2/3,0)`

for the y axis intercepts, simply set x = 0 and solve:
`(3(0)-2)/(0+3) = -2/3`

So the y axis intercept is `(0,-2/3)`
(see the graph of the function below - it's usually good to have this as a "sanity check").

c)f'(x) for this function is found via the rule for finding the derivative of the quotient of two functions.

If `f(x) = (u(x))/(v(x))`, then `f'(x) = (u'(x)v(x) - u(x)v'(x))/(v(x)^2)`

For this function, u(x) = 3x-2, v(x) = x + 3, u'(x) = 3, v'(x) = 1

...put all this together and you get `(3(x+3) - (3x-2))/(x+3)^2`
`=(3x + 9 - 3x - (-2))/(x+3)^2` = `11/(x+3)^2`

...I don't remember how to do a "sign table" for this, but it shouldn't be a big deal: function f'(x) is always positive ( greater than 0)

d) stationary points: not an issue, I don't think, because this function doesn't have any. f'(x) does not equal zero at any point.

graph{(3x -2)/(x+3) [-20, 20, -10, 10]}

GOOD LUCK