Asked by Unknown
a graph with the points (0,-3) (2,-4) (-2,-4)
What would be the domain, range x-intercepts y-intercepts and funtional value..
Help Please?..
What would be the domain, range x-intercepts y-intercepts and funtional value..
Help Please?..
Answers
Answered by
Steve
just given three points, there's no way to tell anything in general about the function. All we know is
the domain includes {-2,0,2}
the range includes {-4,-3}
y-intercept at y = -3
no x-intercept given
there are many functions which contain these three points:
y = -|x/2| - 3
y = 1/2 cos(π/4 x) - 3.5
y = cos(π/2 x) - 3
y = -x^2/4 - 3
y = 1/4 x^4 - x^2 - 3
and so on
the domain includes {-2,0,2}
the range includes {-4,-3}
y-intercept at y = -3
no x-intercept given
there are many functions which contain these three points:
y = -|x/2| - 3
y = 1/2 cos(π/4 x) - 3.5
y = cos(π/2 x) - 3
y = -x^2/4 - 3
y = 1/4 x^4 - x^2 - 3
and so on
Answered by
Unknown
Thanks?..
Same Type of question...
cAn you tell me if these are maybe right?..
1. points of (2,2) (-1,4) (2,6)
Domain:(-4,2,6)
range:(2,2)
xIntercept: none
y-Intercept: none
functional value: x^2+2
Same Type of question...
cAn you tell me if these are maybe right?..
1. points of (2,2) (-1,4) (2,6)
Domain:(-4,2,6)
range:(2,2)
xIntercept: none
y-Intercept: none
functional value: x^2+2
Answered by
Steve
domain: {-1,2}
range: {2,4,6}
domain is the x-coordinate (1st value)
range is the y-coordinate (2nd value)
f(x) is certainly not x^2 + 2
(-1)^2 + 2 = 3, not 4
(2,6) fits, but (2,2) does not.
In fact, since (2,2) and (2,6) contain the same x-value, but different y-values, these pairs are not even a function.
range: {2,4,6}
domain is the x-coordinate (1st value)
range is the y-coordinate (2nd value)
f(x) is certainly not x^2 + 2
(-1)^2 + 2 = 3, not 4
(2,6) fits, but (2,2) does not.
In fact, since (2,2) and (2,6) contain the same x-value, but different y-values, these pairs are not even a function.
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