Question

use the a, b, and c - sliders to graph the function $f(x)=\frac{x - 1}{x^{2}-4x + 3}$ ($a=-1$, $b=-4$, and $c = 3$). note that the denominator factors to $(x - 3)(x - 1)$. complete parts 1 and 2 below.

click here to launch the interactive figure. use the left and right arrow keys to move along a slider as needed.

part 1: what is the equation of the vertical asymptote?

oa. $y = 0$
ob. $y = 3x$
oc. $x = y$
od. $x = 3$

part 2: what are the coordinates of the hole in the graph?

the coordinates are $\square$. (type an ordered pair. simplify your answer.)

Explanation:

Part 1: Find the vertical asymptote

Step1: Simplify the function

First, factor the numerator \(x^{2}-4x + 3=(x - 1)(x - 3)\). The function \(f(x)=\frac{x - 1}{x^{2}-4x + 3}=\frac{x - 1}{(x - 1)(x - 3)}\). After canceling out the common factor \((x - 1)\) (for \(x
eq1\)), the simplified form is \(f(x)=\frac{1}{x - 3}\) (with a hole at \(x = 1\)).
The vertical asymptote of a rational function \(y=\frac{g(x)}{h(x)}\) (in simplified form) occurs when \(h(x)=0\).

Step2: Solve for \(x\)

Set the denominator of the simplified function \(x-3 = 0\). Solving for \(x\) gives \(x = 3\).

Part 2: Find the coordinates of the hole

Step1: Simplify the function

We have \(f(x)=\frac{x - 1}{(x - 1)(x - 3)}\). The hole occurs at the value of \(x\) that makes the original non - simplified denominator and numerator equal to zero. Here \(x=1\) makes \(x - 1=0\) (in both numerator and denominator).

Step2: Substitute \(x = 1\) into the simplified function (after canceling the common factor)

After canceling \((x - 1)\), the simplified function is \(y=\frac{1}{x - 3}\). Substitute \(x = 1\) into \(y=\frac{1}{x - 3}\), we get \(y=\frac{1}{1-3}=-\frac{1}{2}\)

Answer:

Part 1: D. \(x = 3\)
Part 2: \((1,-\frac{1}{2})\)