11-3-11+Homework+Ditto+expl.

Today in class we worked on the incredibly long homework problem that was assigned the night before. The original problem looked like this: math f(x)=\frac{20x^6-172x^5-607x^4+3446x^3-1711x^2-520x+300}{125x^4-225x^3-390x^2-172x-24} math

The assignment also gave us the factored form that appeared as this:

math f(x)=\frac{(2x-1)^2(5x+2)(x-3)(x-10)(x+5)}{(x-3)(5x+2)^3} math

We then proceeded to cancel the in both the numerator and denominator, as well as the (5x+2) in the numerator and and one (5x+2) in the denominator giving us the following:

math f(x)=\frac{(2x-1)^2(x-10)(x+5)}{(5x+2)^2} where x\neq 3, \frac{-2}{5} math

From this work we deduced the following: Zeros: math \frac{-1}{2}, 10, -5 math

Removable discontinuity: x=3. This is because it media type="custom" key="11229750" is a factor of bot the numerator and denominator therefore that means there is "hole" in the graph at this point.

Vertical Asymptote: media type="custom" key="11229756" this is because media type="custom" key="11229764" appears with a higher degree in the denominator than in the numerator.

Y-intercept: media type="custom" key="11233074"

End Behavior asymptote:

math f(x)=\frac{20x^6-172x^5-607x^4+3446x^3-1711x^2-520x+300}{125x^4-225x^3-390x^2-172x-24} math

We then took did this:

math f(x)=\frac{x^4(20x^2-172x-607+\frac{3446}{x}-\frac{1711}{x^2}-\frac{520}{x^3}+\frac{300}{x^4})}{x^4(125-\frac{225}{x}-\frac{390}{x^2}-\frac{172}{x^3}-\frac{24}{x^4})} math

Which led to this:

math f(x)=\frac{20x^2-172x-607+\frac{3446}{x}-\frac{1711}{x^2}-\frac{520}{x^3}+\frac{300}{x^4}}{125-\frac{225}{x}-\frac{390}{x^2}-\frac{172}{x^3}-\frac{24}{x^4}} math


 * Except that Mrs. Tyson was hoodwinked. We do need to do the long division... I'll insert it as a .gif later.**

The following information can be found from the previous equation:

math x \to \infty, f(x)=\frac{20x^2-172x-607}{125} math

Using ALL of the above information we get a graph that looks like this:

To find the tangent line assume the following:

math f(x)=A+B(x-8)+C(x)(x-8)^2 math

Plug in 8 to get A

math \frac{-325}{98}=f(8)=A+0 math

Plug back in A to the other side:

math f(x)+\frac{325}{98}=B(x-8)+C(x)(x-8)^2 math

TIME TO SOLVE FOR B!

math \frac{num(f(x))\cdot 98+325(den(x))}{98(denf(x))}=\frac{1960x^6-16856x^5-18,861x^4+264,583x^3-294,498x^2-106,860x+21,600}{98(denf(x))} math

Synthetic division to follow:

__8|__ 1960 -16,846 -18,861 264,583 -294,428 -106,860 21,600 __15,680 -9,408 -226,152 307,448 104,160 -21,600__ 1960 -1,176 -28,269 38,431 13,020 -2,700 0

This is all we completed in class, however from here you would write this out and it is the numerator. Then put it over 98 times the denominator of f(x) plug in 8 and solve for B!

math f(x)=B+C(x)(x-8)=\frac{1960x^5-1,176x^4-28,269x^3+38,431x^2+13,020x-2,700}{98(125x^4-225x^3-390x^2-172x-24)}=\frac{1960x^5-1,176x^4-28,269x^3+38,431x^2+13,020x-2,700}{12,250x^4-22,050x^3-38,220x^2-16,856x-2352} math

Then plug in 8 to solve for B :-)