a. 6*(x-1)(x+1)

^{--------------}

(x-2)(x-5)

b. x not equal to 2 or 5, x∈R

c. at x=2 and x=5

d. at x→∞, f(x)→6

e.x=0 f(0)=-3/5

f.f(x)=0 at x=±1

Jim

Theresa L.

asked • 04/23/14Let f(x)= 6x^2-6

x^2-7x+10

(a) Write f(x) with the numerator and denominator completely factored.

(b) State the domain.

(c) State the vertical asymptote(s).

(d) State the horizontal asymptote.

(e) State the y-intercept.

(f) State the x-intercept(s).

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a. 6*(x-1)(x+1)

(x-2)(x-5)

b. x not equal to 2 or 5, x∈R

c. at x=2 and x=5

d. at x→∞, f(x)→6

e.x=0 f(0)=-3/5

f.f(x)=0 at x=±1

Jim

Philip P. answered • 04/23/14

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f(x) = (6x^{2}-6)/(x^{2}-7+10)

a) Write f(x) with the numerator and denominator completely factored

(6x2-6) factors to 6(x^{2}-1) = 6(x+1)(x-1)

x^{2}-7+10 factors to (x-5)(x-2)

You might want to check out this tutorial on factoring to learn how to do it for yourself.

(b) State the domain

Division by zero is undefined, so the denominator cannot = 0. It equals zero when x=5 or x=2. Hence the domain is all real numbers except 2 and 5.

(-∞,2)U(2,5)U(5,+∞) [interval notation]

(c) State the vertical asymptote(s)

Vertical asymptotes exist where the denominator goes to 0. Hence we have one at x=2 and another at x=5.

(d) State the horizontal asymptote.

Horizontal asymptotes occur when x approaches ±∞. As x gets very large or small (x+1), (x-1), (x-5), and (x-2) all approach x. Picture x as 10.000.000. The difference between 99,999,995 (x-5) and 10,000,001 (x+1) is very small. Hence the x terms in the numerator and denominator cancel leaving only the 6. So there is a horizontal asymptote at y=6.

(e) State the y-intercept.

The y-intercept occurs when x=0. Plug zero in for x in the factored equation for f(x) in step (a) to compute the y-intercept.

(f) State the x-intercept(s)

The x-intercepts occur when y=0. This can only happen when the numerator of f(x) = 0. What values of x will cause the numerator of f(x) to be zero? There are two answers. Use the factored form of f(x) in (a).

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