From 240912d7ad79bebb20724c506b0040ee1dbb1042 Mon Sep 17 00:00:00 2001 From: kathypinzon <75687638+kathypinzon@users.noreply.github.com> Date: Sun, 28 Jul 2024 22:58:22 +0000 Subject: [PATCH] almost all answers to PR6 --- source/04-PR/06.ptx | 244 ++++++++++++++++++++++++++++++++++++++++++-- 1 file changed, 236 insertions(+), 8 deletions(-) diff --git a/source/04-PR/06.ptx b/source/04-PR/06.ptx index be4d5817..36cdc01c 100644 --- a/source/04-PR/06.ptx +++ b/source/04-PR/06.ptx @@ -40,12 +40,24 @@

Find r(1) , r(2) , r(3), and r(4).

+ +

+ r(1) and r(3) do not exist. + r(2)=0 + r(4)=2 +

+
-

Label each of these four points as giving us information about the DOMAIN of r(x), information about the ZEROES of r(x), or NEITHER. +

Label each of these four values as giving us information about the DOMAIN of r(x), information about the ZEROES of r(x), or NEITHER.

+ +

+ r(1) and r(3) give information about the domain. r(2) gives information about the zeroes. r(4) does not give information about either. +

+
@@ -74,6 +86,11 @@

Rewrite r(x) by factoring the numerator and denominator, but do not try to simplify any further. What do you notice about the relationship between the values that are not in the domain and how the function is now written?

+ +

+ r(x)=\dfrac{(x-2)(x-1)}{(x-3)(x-1)} The values not in the domain are zeros of the denominator. +

+
@@ -113,6 +130,50 @@

+ +

+ + + + x + r(x) + + + 2 + 0 + + + 2.9 + -9 + + + 2.99 + -99 + + + 2.999 + -999 + + + 3 + undefined + + + 3.001 + 1001 + + + 3.01 + 101 + + + 3.1 + 11 + + + +

+
@@ -125,6 +186,11 @@
  • As x \to 3 from the right, r(x)\to -\infty
  • + +

    + C and D +

    +
    @@ -163,6 +229,50 @@

    + +

    + + + + x + r(x) + + + 0 + 0.667 + + + 0.9 + 0.524 + + + 0.99 + 0.502 + + + 0.999 + 0.500 + + + 1 + undefined + + + 1.001 + 0.499 + + + 1.01 + 0.497 + + + 1.1 + 0.474 + + + +

    +
    @@ -175,11 +285,21 @@
  • As x \to 1 from the right, r(x)\to -\infty
  • + +

    + A +

    +

    The function is behaving differently near x=1 than it is near x=3. Can you see anything in the factored form of r(x) that may help you account for the difference?

    + +

    + The x-1 factor occurs in the numerator and denominator. +

    +
    +

    Find any holes on the graph of f(x).

    - +

    Sketch the graph of f(x).

    - + +

    + + f(x) = (-(x-1)*(x-4))/(2*(x-1)*(x+3)^2) + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((-3,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,0),(x,-7,7),color='red',linestyle="--") + p + +

    +
    @@ -749,6 +967,11 @@ https://activecalculus.org/prelude/sec-poly-rational-features.html#ez-poly-ratio \displaystyle r(x) = \frac{x^2-16}{x+4}

    + +

    + At x=-4 there is a hole since the function values tend towards a number. +

    +
    @@ -756,6 +979,11 @@ https://activecalculus.org/prelude/sec-poly-rational-features.html#ez-poly-ratio \displaystyle s(x) = \frac{(x-2)^2(x+3)}{x^2 - 5x - 6}

    + +

    + There are no holes. +

    +