At which values of x does the function F(x) have a vertical asymptote? Check all that apply.F(x) = 17(𝑥+2)(𝑥+3)7(x+2)(x+3)1 A.7B.-2C.3D.-3E.2SUBMITarrow_backPREVIOUSNEXTarrow_forward
Question
At which values of x does the function F(x) have a vertical asymptote? Check all that apply.F(x) = 17(𝑥+2)(𝑥+3)7(x+2)(x+3)1 A.7B.-2C.3D.-3E.2SUBMITarrow_backPREVIOUSNEXTarrow_forward
Solution
The function F(x) = 17/(x+2)(x+3) has vertical asymptotes where the denominator of the function is equal to zero. This is because division by zero is undefined in mathematics.
To find the vertical asymptotes, we set the denominator equal to zero and solve for x:
(x+2)(x+3) = 0
Setting each factor equal to zero gives the solutions x = -2 and x = -3.
So, the function F(x) has vertical asymptotes at x = -2 and x = -3. Therefore, the correct answers are B. -2 and D. -3.
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