Question by E: Algebra and absolute value function help?
Describe how the variables a, h, and k from the vertex form of an absolute value function , affect the transformation for the parent function y=|x|. Be sure to use math language, include graphs, t-tables, and algebraic examples.
Can someone please help me with this? I have to write a whole report answering this! It’s due tomorrow (Monday, January 27).
Oops sorry I mean Monday January 28. Today’s the 27th.
Best answer:
Answer by p lawes
You did not include a, h, or k in your problem, but h and k are usually with x and y respectively, and I am guessing the a is in front of the absolute value, or the f(x) symbol. Something like:
y = a|x- h| +k Right?
Maybe f(x) = a|x- h| +k
Ok,
The “h” shifts, or tranlates horizontally, the function in the OPPOSITE x direction. So |x + 2| would shift the graph 2 places to the LEFT, or the negative direction. Likewise, |x – 4| would shift it 4 place to the RIGHT, or the positive direction.
The “k” shift it up or down, but this time in the same direction.
+3 shift up, and -3 shifts down.
The “a” is a scalar and it is multiplied times the absolute value. Notice it is outside the function so we are scaling the “y” value. We call what the “a” does “scaling”. It does exactly what it says it does. 2|x| multiplies the y by 2.
Examples: y = |x + 3| shifts the x 3 places in the negative direction: to the left.
y = |x| + 3 shifts the y 3 places in the positive direction: upwards
y = 4|x| scales the y by 4. It expands it upwards 4 units.
What about combinations?
y = 3|x-1| – 4.
We know the 4 shifts it down 4, the -1 shifts it 1 to the RIGHT, and the 3 stretches the y number by 3.
But do I stretch first or shift down 4 first? For the y value, follow the order of operations. PEMDAS. Multiply the 3 then subtract the 4
*********************************************************************
as far as T-Tables, just make a function using numbers for h, k, and a. Then make a column for x and a column for y.
plug in the x and calculate the y!
Here’s a link that might help:
http://www.richland.edu/james/lecture/m116/functions/translations.html
**********************************************************************
Now for the Algebra; “Solve for the x, Substitute for the y”
Given a point on f(x), lets say (-5, 5), how can you find point on the new function, say y = -2|x + 3| – 4?
In the original function, your x number was -5 and your y number was 5.
In the equation, what’s in the absolute value is the x, and the absolute value itself is the y:
x+3 is the x, and |x+3| is the y
********* “Solve for x, Substitute for y” ********
“Solve for x” y = -2|x + 3| – 4
Your x numer is -5 so solve x+3 = -5. You get x = -8. You shifted 3 places to the left.
“Substitute for y” y = -2|x + 3| – 4
Your y number was 5, so -2(5) – 4 = -14
Your new point is (-8, -14). You can check by putting -8 in for x and see if you get -14 for y. (You do!)
For the point (0, 0) what would the new point be?
Solve for x: x + 3 = 0 (The x number), so x = -3
Substitute for y: -2(0) – 4 = -4, so y = -4
(-3, -4)
plug in the -3 for x and check that you get the -4 for y.
For the point (-1,1), what would the new point be?
Solve for x: x+3 = -1, x = -4
Substitute for y: -2(1) – 4 = -6, y = -6
(-4, -6)
It works for any function. In y = x^2 we have the point (2, 4)
Find the corresponding point be on y = 2(x-1)^2 + 3
The thing being squared is the x, and what I get after I square it is the y:
In the function y = 2(x-1)^2 + 3 the x is the (x-1), and the y is the (x – 1)^2
Solve for x, substitute for y>
x – 1 = 2, x = 3
2(4) + 3, y = 11
(3, 11) double check by putting the 3 in the new function for x, and make sure you get 11 out for y. (You do!)
what would the corresponding point be for (-3, 9)?
Scroll down to see the answer
(-2, 21)
check and see!
Good luck!
What do you think? Answer below!