User talk:Aznshorty67

$$\mbox{Find all asymptotes and extrema (by hand). Sketch a graph.}$$

$$y=\frac{x^2}{x^2-4x+3}$$

$$\mbox{First, see where the function is undefined to find the asymptotes. Set the denominator to zero.}$$

$$x^2-4x+3=0$$

$$\mbox{Solve for } x.$$

$$(x-3)(x-1)=0$$

$$(x-3)=0\ \ \ \ \ (x-1)=0$$

$$x=3\ \ \ \ \ \ \ \ \ \ \ \ \ x=1$$

$$\mbox{Vertical asymptotes are at }x=1 \mbox{ and }x=3$$

$$\mbox{Since the highest powers of both the numerator and the } $$

$$\mbox{denominator are the same, you can divide the coefficients to get the horizontal asymptotes.}$$

$$y=\frac{1x^2}{1x^2-4x+3}$$

$$y=\frac{1}{1}$$

$$\mbox{The horizontal asymptote is at }y=1$$

$$\mbox{To find the extrema, first find the derivative using the quotient rule.}$$

$$\frac{dy}{dx}=\frac{(x^2-4x+3)(2x)-(2x-4)(x^2)}{(x^2-4x+3)^2}$$

$$\mbox{Then find the critical numbers by finding where the derivative is zero or undefined.}$$

$$\mbox{First set the numerator to zero to find where the derivative is zero.}$$$$(x^2-4x+3)(2x)-(2x-4)(x^2)=0$$

$$\mbox{Distribute and simplify.}$$

$$2x^3-8x^2+6x-2x^3+4x^2=0$$

$$-4x^2+6x=0$$

$$\mbox{Solve for } x.$$

$$x(-4x+6)$$

$$x=0 \ \ \ \ \ x=\frac{3}{2}$$

$$\mbox{Now set the denominator to zero to find where the derivative is undefined.}$$

$$(x^2-4x+3)^2=0$$

$$(x-1)(x-3)=0$$

$$x=1\ \ \ \ \ \ x=3$$

$$\mbox{The critical numbers are}$$

$$x=0, 1, \frac {3}{2}, 3$$

$$\mbox{Test if the critical numbers are extrema using the first derivitave test. }$$

$$x=0 \mbox{ is a local minimum because }f'(x) \mbox{ changes from negative to positive at } x=0$$

$$x=\frac{3}{2} \mbox{ is a local maximum because }f'(x) \mbox{ changes from positive to negative at } x= \frac{3}{2}$$

$$\mathbf{Ken Kuo, Sergio Leal, Patrick Lu, Steven Maheshwary}$$