Sketch the graph indicating the horizontal and vertical asym

Sketch the graph indicating the horizontal and vertical asymptotes and intercepts. Complete the table, Report answers as equations or ordered pairs as necessary. f(x) = 3x^2 + 5x - 12/x^2 - 2x - 15 Describe the domain, end behavior and what happens to the y-values as x approaches discontinuities for the function from the above problem. Identify each discontinuity as to type. Solve the following using sign analysis. 3x^2 + 5x - 12/x^2 - 2x- 15 greaterthanorequalto 0

Solution

13)

f(x)=(3x²+5x-12)_/(x²-2x-15)

f(x)=((3x-4)(x+3))_{/((x-5)(x+3))} , domain (-,-3)U(-3,5)U(5,)

HA: y=3

VA: (x+3) cancels out

denominator=0

x-5=0

x=5

x intercept :(x+3) cancels out

numerator=0

3x-4=0

x=4_/3

point is (4_/3,0)

y intercept:x=0

y=(0+0-12)_/(0-0-15)

y=4_/5

point is (0,4_/5)

RD means?

---------------------------------

14), domain (-,-3)U(-3,5)U(5,)

end behaviour:

lim_[x->]f(x)

=lim_[x->](3x²+5x-12)_/(x²-2x-15)

=3

lim_[x->-]f(x)

=lim_[x->-](3x²+5x-12)_/(x²-2x-15)

=3

at discontinuities:

at x =-3

lim_x->-3 f(x)

=lim_x->-3((3x-4)(x+3))_{/((x-5)(x+3))}

=lim_x->-3(3x-4)_/(x-5)

=(-9-4)_/(-3-5)

=13_/8

lim_x->5^- f(x)

=lim_x->5^-((3x-4)(x+3))_{/((x-5)(x+3))}

=lim_x->5^-(3x-4)_/(x-5)

=-

lim_x->5⁺ f(x)

=lim_x->5⁺((3x-4)(x+3))_{/((x-5)(x+3))}

=lim_x->5⁺(3x-4)_/(x-5)

=

removable discontinuity at x =-3

infinate discontinuity at x =5

-----------------------------------

15)

(3x²+5x-12)_/(x²-2x-15)=((3x-4)(x+3))_{/((x-5)(x+3))}

(3x²+5x-12)_/(x²-2x-15)=((1_/3)(x-(4_/3))(x+3))_{/((x-5)(x+3))}

x...........................-3.......................4_/3.....................5...................

f(x).............+........dne........+............0............-........dne.........+...................

(3x²+5x-12)_/(x²-2x-15)>=0

x=(-,-3)U(-3,4_/3]U(5,)