I'm not really sure how to even start this question. Any help would be much appreciated. Thank you
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Horizontal asymptotes occur when lim x→∞ f(x) = L, where y=L is the equation of the asymptote. Vertical asymptotes occur in rational functions where the denominator equals zero at a certain x value. Even functions are symmetric across the y-axis, and usually have an even number as the highest degree.
lim x→∞ f(x) = 1
something / x has a vertical asymptote at x=0.
In order for f(x) to be even, the degree in the numerator will have to be 2 or higher, but to have this condition, the denominator's degree must also be even.
f(x) = x²/x²
This can't be right, though, because f would simplify to 1, which has no vertical asymptote.
You can make it right by adding a constant to the numerator. For example f(x) = (x² - 1)/x² would work.
lim x→∞ f(x) = 1
something / x has a vertical asymptote at x=0.
In order for f(x) to be even, the degree in the numerator will have to be 2 or higher, but to have this condition, the denominator's degree must also be even.
f(x) = x²/x²
This can't be right, though, because f would simplify to 1, which has no vertical asymptote.
You can make it right by adding a constant to the numerator. For example f(x) = (x² - 1)/x² would work.
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Hello,
An even function is a function that verify the relationship f(x)=f(-x)
I propose:
f(x) = √(x² + 1) / |x|
This function is even since:
f(x) = √(x² + 1) / |x| = √[(-x)² + 1] / |-x| = f(-x)
There a vertical asymptote x=0 and
Lim (x→±∞) f(x) = 1 means there is an horizontal asymptote y=1.
Regards,
Dragon.Jade :-)
An even function is a function that verify the relationship f(x)=f(-x)
I propose:
f(x) = √(x² + 1) / |x|
This function is even since:
f(x) = √(x² + 1) / |x| = √[(-x)² + 1] / |-x| = f(-x)
There a vertical asymptote x=0 and
Lim (x→±∞) f(x) = 1 means there is an horizontal asymptote y=1.
Regards,
Dragon.Jade :-)