Question

H(x)= (x²+8x+16)/x²+6x+8 At each of the following values of x, select whether h has a zero, a vertical asymptote, or a removable discontinuity.

Zero Vertical Asymptote Removable Discontinuity
x=-4 O O O
x=-2 O O O

Answer 1

After factoring, h(x) simplifies to h(x) = (x+4)/(x+2), resulting in a removable discontinuity at x = -4 and a vertical asymptote at x = -2. There are no zeros for these values of x.

Step-by-step explanation:

To determine whether the function h(x)= (x²+8x+16)/(x²+6x+8) has a zero, a vertical asymptote, or a removable discontinuity at the given values of x, we need to factor both the numerator and the denominator, if possible, and then evaluate the function at the given x values.

First, factor the numerator and the denominator:

• Numerator: x²+8x+16 can be factored into (x+4)².
• Denominator: x²+6x+8 can be factored into (x+4)(x+2).

Now the function can be simplified to h(x) = (x+4)/(x+2), after canceling out the common factor of (x+4) from the numerator and the denominator.

Let's analyze the function at the given x values:

• x = -4: When x equals -4, this inserts the common factor of (x+4) that we canceled out earlier. Because the cancellation removed what would have been a zero in both the numerator and denominator, there is now a removable discontinuity at x = -4.
• x = -2: When x equals -2, the denominator will be zero (since x+2 = -2+2 = 0), which means there is a vertical asymptote at x = -2.

Therefore, h has a removable discontinuity at x = -4 and a vertical asymptote at x = -2. There is no zero at either x = -4 or x = -2.