The asymptote cannot be x= because x can be any number. If you think about it, you can take a number to any exponent.
If x is a positive exponent, y is positive.
If x is a nevative exponent, y decreases, but is still positive. This is because a number to a negative exponent equals 1 over the number to the positive exponent. Thus, it is smaller, but still positive.
If x is 0, y is positive again because anything to the zero is positive 1.
There is no way y could be less than or equal to zero. So, there is an asymptote at y=0.
Also, set the equation equal to 0 and solve. You should end up with 4^x=0. Since no exponenent can make a number zero, this isn't possible, so y cannot equal zero.
Here is the graph for a visual:
Answer:
(x-3)^2 + (y+2)^2 = 9^2
Step-by-step explanation:
x^2 -6x+y^2+4y-68 = 0
Complete the square
x^2 -6x+y^2+4y-68+68 = 0+68
x^2 -6x+y^2+4y = 68
Find the term to add for x
-6 /2 = -3 -3^2 = 9
Find the term to add for y
4/2 =2 2^2 = 4
Add 9 and 4
x^2 -6x+9+y^2+4y+4 = 68+9+4
(x-3)^2 + (y+2)^2 = 81
(x-3)^2 + (y+2)^2 = 9^2
The standard form is
(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius
1st. Try to make x by itself. Divide 12 on each side.
After it will look like this
x=12
Answer:
Step-by-step explanation:
Note that if it has a y- intercept of 20, this means that when x = 0 , y = 20
Find the table which shows that :)