Use the slope formula or since it's a straight line crossing the y axis at one, it means the y=1
The derivitive of sec(x) is sec(x)tan(x)
find the slope at pi/3
sec(pi/3)=2
tan(pi/3)=√3
sec(x)tan(x) at x=pi/3 is 2√3
for slope=m and a point is (x1,y1)
the equation is
y-y1=m(x-x1)
slope=2√3
point=(pi/3,2)
equation is
y-2=2√3(x-pi/3)
Answer:
f(7)=13
g(-9)=86
Step-by-step explanation:
f(7)=2*7-1=14-1=13
g(-9)=(-9)^2+5=81+5=86
A VA is a vertical line that an equation will approach but never touch.
To solve for a VA take the denominator and set it equal to zero because a number divided by zero is undefined. (You can check this on a calculator by taking 2/0, you will get error)
For this equation set x+4=0, solve for x, you will get x=-4 as your VA
A HA is similar since it is a horizontal line that an equation approaches but never touches.
To solve for an HA I learned a rule called the BB rule or B00bie Butt rule. I know it is silly but it is easy to remember. Here are three examples:
(1) 2x^2/x^2
(2) 2x/x^2
(3)2x^2/x
To solve you look at the degree of the equation.
(1) This has the same degree on top and bottom (x^2) so you take the leading coefficient (2/1). HA=2
(2) This has a higher degree on the denominator (butt) so it is a big honking zero. HA=0
(3) This has a larger degree on top (b00bie) so it has no asymptote. Again it's silly but I think if the "b00bies" are too big they are gonna fall over and isn't going to work. HA=DNE
For yours you have the degree x/x which are the same so the leading coefficiant (numbers in front of variable) equals 1. HA=1
Holes are points on the equation that are undefined. Yours doesn't have one but on a problem like:
[(x+3)(x-3)]/(x-3) you can cross out the (x-3) on top and bottom which makes a hole. What ever you take out, solve for x when equal to 0:
x-3=0
x=3
Hole at x=3
I know it is wordy but I hope this helps!
