I need help finding the vertical asymptote of
f(x)= 1-x
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2x^2-5x-3
The denominator factors to (2x+1)(x-3)
so it is zero when...x=3, and ...
I still don't exactly know what a vertical asymptote is.
http://www.purplemath.com/modules/asymtote.htm
Thanks that helped alot!
You're welcome.
To find the vertical asymptotes of a rational function, we need to determine the values of x that make the denominator of the function equal to zero.
In this case, the denominator is 2x^2 - 5x - 3. We can try factoring the denominator to solve for x.
To factor, we are looking for two numbers that multiply to give -6 and add to give -5. The numbers that satisfy these conditions are -6 and +1.
So, we can rewrite the denominator as (2x + 1)(x - 3).
Now, we set each factor equal to zero to find the values of x that make the denominator equal to zero:
2x + 1 = 0 => 2x = -1 => x = -1/2
x - 3 = 0 => x = 3
Therefore, the values of x that make the denominator equal to zero are x = -1/2 and x = 3. These are the vertical asymptotes of the function.