Emily
Lv 4
Emily asked in Education & ReferenceHomework Help · 1 decade ago

Absolute-Value Functions?

I'm in desperate need of some homework help. As the title suggests I'm working on absolute value functions, and the teacher didn't explain everything about them...

It's asking for the range of each function, and when it comes to certain functions I'm clueless.

How do I find the range of something like this: y = |6x|.I don't understand this one because I don't know what to do with the number that is in with the x.

Then there's these: y = 6|x| and y = -6|x|. I don't understand what to do with the number on the outside.

And then it goes to things like this: y = |x+6| and y = -|x+6| and y = |6x + 6|

Basically when there's a 6 at any time, I don't know what to do with it! The only answers I did know were things like y = |x| where the range was simply y (is > or = to) 0.

Please help!

2 Answers

Relevance
  • Anonymous
    1 decade ago
    Favorite Answer

    The thing to remember about absolute value functions is that they are cheats. They look like a single function, but there are actually two functions there.

    Consider the functional definition of absolute value:

    |x| = x if x >= 0

    = -x if x < 0

    So when you're dealing with an absolute value, you're always going to have to deal with two different parts.

    Take the basic function:

    y = |x|

    When x is positive - on the right side of the y-axis - you're dealing with the line y = x. When x is negative - on the left side of the y-axis - it's the line y = -x. If you graph those, you'll get a V-shape with the point on the origin.

    If you make it y = 6|x|, you're going to have the same general shape. On the right side of the y-axis it'll be the line y = 6x. On the left side, it'll be y = -6x. So you'll have the same V-shape with the point still on the origin, but the slope will be steeper. So - y = a|x| or y = |ax|, you're going to take the basic shape and change the slope.

    If you do something like y = |x| + 6, you're going to have the same shape as y = |x| except it'll be moved up the y-axis a distance of 6. It'll be the same V-shape, but the point will be at (0, 6). Generally, the line y = |x| + c will be the V-shape c units along the y-axis.

    If you do something like y = |x + 6|, that will end up moving it along the x-axis. Consider: the breakpoint for an absolute value is when what's in the absolute value signs is 0. That will happen now when x = -6. So, this will be the V-shape moved six units to the left. y = |x - 6| will be the V-shape moved 6 units to the right.

    Now, put them all together:

    y = |a(x + b)| + c

    That will be the V-shape with a slope of a, moved b units along the x-axis and c units along the y-axis. For instance,

    y = |3(x - 2)| + 7

    will be the V-shape with a slope of 3. The point of the V will be at (2, 7).

    Note the parentheses. If we had y = |3x - 2| + 7, that would still have a slope of 3, but the point would be at (2/3, 7).

    And, a negative in front of the absolute value sign turns the V upside down.

    This is a lot easier to do with a blackboard where you can do sketches. I hope this helps.

  • lavzon
    Lv 4
    4 years ago

    Equations related to absolute values might properly be no longer ordinary to sparkling up. whether, it facilitates to isolate the expression interior easily the cost on one area and then harm the effect into 2 equations, looking on whether the expression interior easily the cost is below 0 or no longer. in this occasion, if we assume that the expression interior easily the cost is destructive, we'd replace easily the cost image with a series of parentheses and multiply it by skill of -one million to indicate the substitute in sign. Or we'd basically stick the destructive sign outdoors the parentheses particularly, which skill a similar factor. Then we'd sparkling up the ensuing equation, additionally determining which values of y (and x) this actual equation applies: x = (-one million)(y + 5); given y + 5 < 0 If y + 5 < 0, then y < -5 and x > 0 fixing for y provides y = -x - 5 We might in addition cope with the case the place the expression interior easily the values is concept to be nonnegative, yet this time we basically replace easily the cost sign with parentheses. because of the fact the expression containing easily the cost is already remoted on one area of the equation, the parentheses are redundant and we are in a position to drop them. If y + 5 ? 0, then y ? -5 and x ? 0 and x = y + 5 fixing for y, we get y = x - 5 that's going to be referred to that even nevertheless x might properly be considered a function of y, y won't be in a position of be considered a function of x because of the fact for each cost of x different than x = 0, we get 2 accessible values of y. the answer will for this reason be y = x - 5 OR y = -x - 5 if x ? 0. a similar concerns be conscious while working with inequalities, different than while the inequality image factors in direction of the expression contained in easily the cost image, the equations are coupled with AND particularly than OR. this could be real whether the is a strict one (< or >) or additionally includes equality (? or ?) development an inequality by skill of negating an equation (?) additionally variations the OR logical conjunction to an AND.

Still have questions? Get your answers by asking now.