Which is true for the graph of x^2-9/x^2-4
the answer is vertical asymptotes at x = ±2 but i don't understand why.
look at the denominator. What happens to a function which has a zero in the denominator?
i don't understand what you mean
consider y = 1/x
division by zero is not defined. The closer x gets to 0, the larger y gets. At x=0, y is undefined, infinitely large.
Same for this problem. What happens to y as x gets close to 2? x^2-4 gets close to 0, and you cannot divide by 0.
To understand why the graph of the rational function f(x) = (x^2 - 9) / (x^2 - 4) has vertical asymptotes at x = ±2, we need to analyze the behavior of the function as x approaches these values.
First, let's determine where the function is undefined. In this case, the function is undefined when the denominator becomes zero, which occurs when x^2 - 4 = 0. Solving this equation, we find that x = ±2.
Now, we'll examine the limit of the function as x approaches these values.
When x approaches 2, the expression (x^2 - 9) / (x^2 - 4) can be simplified by factoring as follows:
(x^2 - 9) / (x^2 - 4) = [(x - 3)(x + 3)] / [(x - 2)(x + 2)].
As x gets larger and approaches 2 from the right side (x > 2), both the numerator and denominator approach zero, since (x - 3) and (x + 3) approach zero. However, the ratio of these two expressions [(x - 3)(x + 3)] / [(x - 2)(x + 2)] gets infinitely larger, so the limit approaches positive infinity. Consequently, the graph has a vertical asymptote at x = 2.
Similarly, when x approaches -2, the expression (x^2 - 9) / (x^2 - 4) can be factored as:
(x^2 - 9) / (x^2 - 4) = [(x - 3)(x + 3)] / [(x - 2)(x + 2)].
As x approaches -2 from the left side (x < -2), both the numerator and denominator approach zero. However, the ratio [(x - 3)(x + 3)] / [(x - 2)(x + 2)] gets infinitely larger but negative. Thus, the limit approaches negative infinity. Accordingly, the graph has a vertical asymptote at x = -2.
In summary, the graph of the function f(x) = (x^2 - 9) / (x^2 - 4) has vertical asymptotes at x = ±2 because the limits of the function approach infinity and negative infinity respectively as x approaches these values.