Lets say we have
P(x)/q(x)
vertical assymtotes are in the form x=something, not y=0
y=0 are horizontal assemtotes
so verticall assymtotes
reduce the fraction
set the denomenator equal to zero
those values that make the deomenator zero are the vertical assymtotes
the horizontal assymtote
when the degree of P(x)<q(x), then HA=0
when the degree of P(x)=q(x), then divide the leading coefient of P(x) by the leading coeficnet of q(x)
example, f(x)=(2x^2-3x+3)/(9x^2-93x+993), then HA is 2/9
ok so for vertical assymtote example
f(x)=x/(x^2+5x+6)
the VA's are at x=-3 and x=-2
horizontal assymtote
make degree same
f(x)=(3x^2-4)/(8x^2+9x),
the HA is 3/8
hope I helped, read the whole thing then ask eusiton
Y=-2x-2 it is the only one that is perfectly perpendicular and passes through (-3,4)
<span>Assume that,
x^2+4x-5 = 0 .......(1)
Then,
x^2+4x-5 = 0
x^2+5x-1x-5 =0
x(x+5)-1(x+5) = 0
(x+5) (x-1) = 0
We get x=-5 and x=1
Sub x=-5 in equ (1)
(-5)^2+4(5)-5 = 0
-25+20-5 = 0
-25+25= 0
0 = 0
Sub x=1 in equ (1)
(1)^2+4(1)-5 = 0
1+4-5 = 0
5-5 = 0
0 = 0
Therefore x value is -5 and 1</span>
Answer:
simplified equation is 10y-10x
Step-by-step explanation:
8(y-x) -2(x-y)
multiple 8 and 2 by the corresponding parentheses
8y-8x -2x +2y
10y -10x
Answer:
x = -2
Step-by-step explanation:
The tangent line has length "x + 8"
The secant line has length "x + 6 + 5", where
5 is the inner part
x + 6 is the outer part
Now,
the secant-tangent theorem tells us that square of the tangent line is equal to the outer segment of secant line multiplied by length of whole secant line.
So, we can say:
We can solve for x shown below:
The value of x is -2
