Monday, October 8, 2007

Graphing |f(x)|

My apologies for the very late scribe post.

So, on Thursday, we have been given a surprise. We had a substitute which was Ms. Nickelson, BUT, that was not the surprise. She had been given us a POP QUIZ on Transformations, using NO calculators. It all turned out very well than we thought. The quiz was quite easy once after we finished.

We then, were given a new lesson to be learned; Graphing f(x).

Graphing Absolute Value Functions

We were given y = x to start off. On the graph above it shows a purple line which indicates y = x. The question then was added by Absolute Values; y = x. The blue line on the graph above indicates y = x. How did i get that V shape looking graph you may ask? Well because of those absolute values in front and after the x, you make the x values that are negative, positive. once you've done that, you'll end up having a V shape graph, and there you have it, you've graphed an absolute value function.

Example #1

we first graphed y = 3x - 4, which is the purple line that is on the graph above.

To find out where the y intercept is, we made x equal zero. there fore, the y intercept is at (0, -4). after we plotted the dot, we would easily find the rest of the dots and connect them by finding out what the slope is. y = mx + b. the m indicates the slope. rise over run. so starting from the point (0, -4) we rise 3 then run 1. once after that is all graphed, it should look like the purple line above.

we were given a second function to graph which is y = 3x - 4, which is the blue line above. what we did to get our original line to that function, was making all the negative values become positive values. then after we factor out the three;

y = 3 x-4/3

we shift 4/3 to the right, and the 3 indicates a horizontal compression. the y intercept is at 0 and the x intercept is at 4/3. once that all is done, it should look like the blue graph i have up above. the domain and range for this absolute graph is D: XER or (-00, 00), R: [0, 00).

Example #2

the blue line indicates y = -x +4. i got that line by finding out that the x and y intercepts were which was (o, 4) and (-4, 0).

we then were given to graph the absolute value of that line. y = -x +4;

first off, we make all negatives become positives and factor out the negative.

y = -1 x+4

it would still look the same because the absolute negative 1 does not affect the graph since it is an absolute value, which can't be negative.

D: (-00, 00)

R: [0, 00)

Zeroes at; X = 4

Example #3

The black absolute function above on the graph is the original function of y = x. we were given to graph 4 x + 2 - 5.

4 = compress

finding where Zero's are;

0 = 4 x+2 - 5

5/4 = 4 (x+2/4) --> (the 4's reduce)

5/4 = x+2

5/4 = x+2 OR -5/4 = x+2

5/4 - 8/4 = x OR -5/4 - 8/4 = x

-3/4 = x OR -13/4 = x

Therefore, Zeros; X = -3/4 , and -13/4.

D: (-00, 00)

R: [-5, 00)

Example #4

we where given to graph y = x^2 - 16 and y = x^2 - 16.
the function y = x^2 - 16 is shown above in black. we already know what shape the function is going to be before graphing it, given by the hint of the x^2 which is obviously a parabola. to find where the zero's are on the x axis, you factor out the y = x^2 - 16. which will become the factors of ;
Zero's; X = -4, and 4.

but when graphing y = x^2 - 16, we make all the negative values become positive. and then you should have an upside down parabola just like the red one is shown up above.

Assignment in class was to do Exercise 11 questions # 1 - 20. unfinished is for homework

"drums please................ Friday's scribe... will be..... Haiyan"

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