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It is y = 0

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Q: What is the horizontal asymptote of y equals 2 to the power of x?
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Which function has the following a vertical asymptote at x equals -4 horizontal asymptote at y equals 0 and a removable discontinuity at x equals 1?

2x-2/x^2+3x-4


What is the horizontal asymptote of y equals x divided by x2 plus 2x plus 1?

y = x / (x^2 + 2x + 1) The horizontal asymptote is y = 0


How do you find the horizontal asymptote in a graph of a rational function?

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.


What is the domain range and asymptote of gx equals 2 to the power of x minus 3?

The domain is (-infinity, infinity) The range is (-3, infinity) and the asymptote is y = -3


What is the horizontal asymptote of an equation where the leading coefficient of the numerator is greater than that of the denominator?

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


What is the formula for a rational function with x intercepts at x equals 3 and x equals negative 2. y intercept at y equals 12. a hole at x equals 1 and no horizontal asymptote?

f(x) = 2*(x-3)*(x+2)/(x-1) for x ≠ 1


What is the domain range asymptote and intercept of the equation y equals 4 times 2 exponent x?

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)


How do you find a horizontal asymptote?

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.


Is y equals 2 to the x power a linear equation?

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.


What is the domain of the function fx equals 3 over x plus 2?

- 2 makes this zero and provides the vertical asymptote. So, from - infinity to - 2 and from - 2 to positive infinity


What is the vertical asymptote of 4 divided by x2?

2


How do I determine the domain and the equation of the vertical asymptote with the equation f of x equals In parentheses x plus 2 end of parentheses minus 1?

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.