F(x)=(2x^2-2x-12)\3X-12 in factored form and reduced form.

Question

I am doing a worksheet on curve sketching, and I need to know the factored form, reduced form, HA or SA, the domain, holes, VA, x-int, and y-int. If you could help me and explain how to do it so I can do the rest, that would be great thanks so much!!

Answer 1

F(x)=(2x²-2x-12)/(3x-12),

F(x)=2(x²-x-6)/(3(x-4))=⅔(x-3)(x+2)/(x-4).

Domain is all x except x=4;

VA (vertical asymptote) at x=4;

x-intercepts: x=3 and -2;

y-intercept at -12/(-12)=1;

SA (sloped asymptote): y=⅔x;

no HA (horizontal asymptote) or holes.

To find SA plug in a very large value for x, so F(x)→⅔x²/x=⅔x.

For example, let x=1000, F(1000)=668.67 approx, and ⅔×1000=666.67. F(x)>⅔x so the curve lies above the asymptote.

When x=-1000, F(-1000)=-664.67 approx, and ⅔(-1000)=-666.67. F(x)>⅔x so the curve still lies above the asymptote. But when x=4-Δ, where Δ is small and positive, F(4-Δ)=⅔(1-Δ)(6-Δ)/(-Δ)→-4/Δ→-∞ as Δ→0, so the curve crosses the asymptote when x<4 and when ⅔(x-3)(x+2)/(x-4)=⅔x, so x²-x-6=x(x-4)=x²-4x, -x-6=-4x, 3x=6, x=2. Between x=2 and 4, the curve dips down to -∞. F(2)=⅔(-1)(4)/(-2)=4/3, which means the intersection occurs at (2,4/3). This is useful to know when plotting the graph. When x=4+Δ (the right side of the VA), F(x)=⅔(1+Δ)(6+Δ)/Δ→4/Δ→+∞ as Δ→0, so the curve doesn't intersect the asymptote at all when x>4: it simply "rides" the SA.

In the graph below the curve is shown in red, the VA in green, and the SA in blue, and the x- and y-intercepts and SA intersection are labelled.