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Paladinen [302]
3 years ago
8

The absolute value function is written f(x) = |x|. It is composed of the lines y = x and y = -x. Its domain is the real numbers

and its range is the set of real numbers: [0, ∞). Its graph is:
1. The absolute value function is y = |x|. Is it possible for the absolute value function to ever have a negative y value? Here is a hint...I’m looking for the actual definition of the absolute value in terms of distance with a detailed example to support your response. Is there a way to transform an absolute value function to have negative outputs?
2. Define the absolute value function, y = |x|, as a piecewise function. Please include complete sentences and examples to justify your answer to receive credit.

Mathematics
1 answer:
Katena32 [7]3 years ago
4 0
1.
no, there will never be a negative y-value.  <span>y= |x| will always be nonnegative. |x| can be distance x is from 0 and a distance can never be negative.

</span>2.
you can define it as
y = |x| = x if x ≥ 0, -x if x < 0

absolute value can be interpreted as a function that does not allow negative real numbers, forcing them to be positive (leaving 0 alone). if the input x is more than or equal 0, then x stays positive so there is no need to do anything: "x if x ≥ 0".
if the input is less than 0, then it is an negative number and needs a negative coefficient to negate the negative: "-x if x < 0"

example: if x = -3, then it will take the "-x if x < 0" piece resulting in y = -(-3) = 3, which is what |-3| does

if x = 1, it will take the "x if x ≥ 0" piece and just have y = 1 which is what |1| does.

for x = 0, it will take the "x if x ≥ 0" and just have y = 0 which is what |0| does
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Point B has coordinates ​(​1,2​). The​ x-coordinate of point A is -8. The distance between point A and point B is 15 units. What
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Answer:

The possible coordinates of point A are A_{1} (x,y) = (-8, 14) and A_{2} (x,y) = (-8, -10), respectively.

Step-by-step explanation:

From Analytical Geometry, we have the Equation of the Distance of a Line Segment between two points:

l_{AB} = \sqrt{(x_{B}-x_{A})^{2} + (y_{B}-y_{A})^{2}} (1)

Where:

l_{AB} - Length of the line segment AB.

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If we know that l_{AB} = 15, x_{A} = -8, x_{B} = 1 and y_{B} = 2, then the possible coordinates of point A is:

\sqrt{(1+8)^{2}+(2-y_{A})^{2}} = 15

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There are two possible solutions:

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