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# Work the Function f(x)=200x-15/40-2x...and g(x)=x^2/x-2...find asymptotes?

1)Work the Function f(x)=200x-15/40-2x

Find the Vertical, horizonal, or slant asymptotes?

and graph

2)Work the function g(x)=x^2/x-2

Find the Vertical, horizontal, or slant asymptotes?

and graph

### 1 Answer

- Wal CLv 61 decade agoFavorite Answer
1. f(x) = 200x - 15/40 - 2x

I assume you mean

f(x) = (200x - 15)/(40 - 2x)

= 5(40x - 3)/[2(20 - x)]

= 5/2 * (40x - 3)/(20 - x)

= 5/2 * (40x - 800 + 797)/(20 - x)

= 5/2 * -40(20 - x)/(20 - x) +

= 100 + 5/2 * 797/(20 - x)

Now we can deal with it, as messy as it looks as before it was 'improper' since the degree of the numerator was the same as the degree of the denomonator and a rational function is 'improper' if degree numerator ≥ degree denominator

Now lets look at 5/2 * 797/(20 - x)

A. When x < 20, 20 - x > 0 so 5/2 * 797/(20 - x) > 0.

Therefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) > 100

AND when x > 20, 20 - x < 0 so 5/2 * 797/(20 - x) < 0.

Therefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) < 100

ie for x > 20, f(x) < 100 and for x < 20, f(x) > 100

B. When x = 20 20 - x = 0 so 5/2 * 797/(20 - x) is undefined.

Therefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) is undefined.

As x → 20 from the left ,5/2 * 797/(20 - x) → +∞

Therefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x))→ +∞

AND as x → 20 from the right, 5/2 * 797/(20 - x) → -∞

Therefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x))→ -∞

ie vertical asympotes at x = 20

C As x → ±∞, 5/2 * 797/(20 - x) → 0

Therefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) → 100

ie horitontal asymptotes at f(x) = 100

D. You now have enough information to sketch this

2. Work the function g(x)=x^2/x-2

I assume you mean g(x)=x^2/(x - 2)

Again this is improper as deg numerator = 2 and deg denominator = 1 (< deg numerator)

So g(x) = x^2/(x - 2)

= (x^2 - 4 + 4)/(x - 2)

= ((x - 2)(x + 2) + 4)/(x - 2)

= x + 2 + 4/(x - 2)

Do steps A. to D as above

A. Check out the signs of the denominator of the fraction

When x < 2, x - 2 < 0 so 4/(x - 2) < 0

Therefore g)x) = x + 2 + 4/(x - 2) {= x^2/(x - 2)} < x + 2

When x > 2, x - 2 > 0 so 4/(x - 2) > 0

Therefore g(x) = x + 2 + 4/(x - 2) {= x^2/(x - 2)} > x + 2

ie for x < 2, g(x) < x + 2 and for x > 2, f(x) > x + 2

B. When x = 2, x - 2 = 0 so 4/(x - 2) is undefined.

Therefore g(x) = x^2/(x - 2) (= x + 2 + 4/(x - 2)) is undefined.

As x → 2 from the left, 4/(x - 2) → -∞

Therefore g(x) = x^2/(x - 2) (= x + 2 + 4/(x - 2)) → -∞

AND as x → 2 from the right, 4/(x - 2) → +∞

Therefore g(x) = x^2/(x - 2) (= x + 2 + 4/(x - 2)) → +∞

ie vertical asymptotes at x = 2

C. As x → ±∞, 4/(x - 2) → 0

Therefore g(x) = x + 2 + 4/(x - 2) {= x^2/(x - 2)} → x + 2

Therefore g(x) is assymptotic to g(x) = x + 2

D. You now have enough information to sketch this

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