For the function f(x) = 1/(x2+14x-42) there are two vertical asymptotes. Determine the x-coordinate of each one
To determine the x-coordinate of the vertical asymptotes of the function f(x) = 1/(x^2+14x-42), we need to identify the values of x for which the denominator becomes zero.
The denominator of the function is x^2 + 14x - 42. To find the x-coordinate of the vertical asymptotes, we solve the equation x^2 + 14x - 42 = 0.
To solve this quadratic equation, we can either use factoring, completing the square, or the quadratic formula.
One approach is to factor the quadratic equation. We need to find two numbers that multiply to give -42 and add up to 14.
The factors of -42 are:
(-1, 42)
(-2, 21)
(-3, 14)
(-6, 7)
Among these pairs, (-6, 7) is the pair that adds up to 14. Therefore, we can factor the quadratic equation as (x - 6)(x + 7) = 0.
Now, we can set each factor equal to zero, and solve for x:
x - 6 = 0 --> x = 6
x + 7 = 0 --> x = -7
Therefore, the x-coordinates of the two vertical asymptotes are x = 6 and x = -7.