### Let’s do all the steps in graphing the function: 2x  6 y 2 x 9 a)Find x and y intercepts b)Find vertical asymptotes c) Find.

```Let’s do all the steps in graphing the function:
2x  6
y 2
x 9
2
a)Find x and y intercepts
b)Find vertical asymptotes
c) Find horizontal asymptotes
d)Find Relative Extrema
e)Find Inflection points
2x  6
y 2
x 9
2
a) Find intercepts
y-intercept, when x = 0
6 2
y

9 3
Crosses the y-axis here
x-intercept, when y = 0
2x  6
0 2
x 9
2
2x  6
0 2
x 9
2
0  2x  6
2
A fraction is 0 when the
numerator is 0
x 3
The function crosses the x-axis at these two
points.
b) Find vertical asymptotes
2x  6
y 2
x 9
2
At x = 3 and -3
Where the denominator
equals 0 and the numerator
does not equal 0
c) Find horizontal asymptotes
2x  6
y 2
x 9
2
The degree of the top equals
the degree of the bottom
(both 2)
The ratio of leading coefficients is 2/1
So y = 2 is the horizontal asymptote
d) Find relative extrema (Max and Mins)
2x  6
y 2
x 9
2
Need the first derivative
( x  9)(4 x)  (2 x  6)(2 x)
y' 
2
2
( x  9)
2
2
This does not exist at x = 3, and -3 but these
are not critical numbers. Why?
Not in the domain.
Vertical asymptotes
Where is this equal to 0?
( x  9)(4 x)  (2 x  6)(2 x)
y' 
2
2
( x  9)
2
2
Just look at the numerator
( x  9)(4x)  (2x  6)(2x)  0
2
2
( x  9)(4x)  (2x  6)(2x)  0
2
2
Factor a 2x
2x[(x  9)(2)  (2x  6)]  0
2
2
Simplify inside the brackets
2 x[12]  0
The only critical number is x = 0
Do the First Derivative test on x = 0
Use the factored form of the derivative:
2 x(12)
y'  2
2
( x  9)
The denominator is always positive
To the left of 0 choose x = -1. The derivative
is positive here
To the right of 0 choose x = 1. The derivative
is negative here
Therefore, a maximum occurs at the point x = 0
Put this back into the original function:
2x  6
y 2
x 9
2
The maximum is at the point (0, 2/3)
d) Find inflection points
2x  6
y 2
x 9
2
Need the 2nd derivative
2 x(12)
 24x
y'  2
 2
2
2
( x  9)
( x  9)
Here is the 1st
( x 2  9) 2 (24)  (24x)[2( x 2  9)(2 x)]
y' ' 
2
4
( x  9)
( x  9) (24)  (24x)[2( x  9)(2 x)]
y' ' 
2
4
( x  9)
2
2
2
Again, we need only look at the numerator
( x  9) (24)  (24x)[2( x  9)(2x)]  0
2
2
2
This can be factored
 24( x  9)[(x  9)  4x ]  0
2
2
2
After factoring:
 24( x  9)[(x  9)  4x ]  0
2
2
2
and collecting terms inside the brackets
 24( x  9)[3x  9]  0
2
Never 0
2
Never 0
Therefore, there are no inflection points
Intercepts
Asymptotes
```