It is y = 0
2 to the 3rd power equals 8, because 2 times 2 times 2 equals 8.
eight eight
36
-125
Pr(Z > -2) = 97.725%
2x-2/x^2+3x-4
y = x / (x^2 + 2x + 1) The horizontal asymptote is y = 0
The horizontal asymptote is what happens when x really large. To start with get rid of all the variables except the ones with the biggest exponents. When x is really large, they are the only ones that will matter. If the remaining exponents are the same, then the ratio of those coefficients tell you where the horizontal asymptote is. For example if you have 2x3/3x3, then the ratio is 2/3 and the asymptote is f(x)=2/3 or y=2/3. If the exponent in the denominator is bigger, than y=0 is the horizontal asymptote. If the exponent in the numerator is bigger, than there is no horizontal asymptote.
The domain is (-infinity, infinity) The range is (-3, infinity) and the asymptote is y = -3
The leading coefficient doesn't come into play unless certain exponent criteria are matched. I believe that to calculate where the horizontal asymptote is you need to concern yourself with the highest exponent and where it is located ie, the horizontal asymptote for y=(3t^2+5t)/(4t^2-3) is y=3/4
f(x) = 2*(x-3)*(x+2)/(x-1) for x ≠1
y = 4(2x) is an exponential function. Domain: (-∞, ∞) Range: (0, ∞) Horizontal asymptote: x-axis or y = 0 The graph cuts the y-axis at (0, 4)
To Find Horizontal Asymptotes: 1) Put equation or function in standard form. 2) Remove everything except the biggest exponents of x found in the numerator and denominator. then set this number to y= and this is the horizontal asymptote Here is an example: f(x) = (2x^2 + 5x - 3)/(x^2 - 2x) We get rid of everything except the biggest exponents of x found in the numerator and denominator. After we do that, the above function becomes: f(x) = 2x^2/x^2. Cancel x^2 in the numerator and denominator and we are left with 2. The horizontal asymptote for is the horizontal line y=2.
If you mean y = 2^x, then no, it is not a linear equation. This is an exponential equation. The graph of this exponential equation would start out near zero on the left-hand side (there is a horizontal asymptote at y = 0) and would gradually increase as you move to the right: overall, it has a curved shaped. If you mean y = 2x, then yes, it is a linear equation.
- 2 makes this zero and provides the vertical asymptote. So, from - infinity to - 2 and from - 2 to positive infinity
2
The domain of the function f(x) = (x + 2)^-1 is whatever you choose it to be, except that the point x = -2 must be excluded. If the domain comes up to, or straddles the point x = -2 then that is the equation of the vertical asymptote. However, if you choose to define the domain as x > 0 (in R), then there is no vertical asymptote.