It is a relation because it relates two variables (y and x), but is not linear because the x has an exponent of 2.
2007-07-06 06:43:10
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answer #1
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answered by Mr. me 2
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because it has a power of 2 in the equation. a linear function only has variables, such as x, with a power to the first degree.
for example, y=16-x is a linear function
2007-07-06 06:49:17
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answer #2
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answered by borninaugust30 3
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A linear funtion will have an exponent value of 1. You have the term x^2 which makes this a second order function and not linear.
2007-07-06 06:43:37
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answer #3
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answered by El Gigante 4
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A linear equation contains only terms to the first power, and constants (like the 16 in your equation). Anything with a different exponent, like 1/x or x^2 or sqrt(x) is non-linear.
2007-07-06 06:43:34
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answer #4
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answered by lithiumdeuteride 7
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It is a function because for each x value there is 1 and only one y value.
It is not linear because linear functions are those written in the form y=ax+b, where a and b are constants and a is not equal to 0. The graph is a straight line. Your graph is a parabola.
It is not a one to one function because there are y values that have more than one x values (in other words, it fails the horizontal line test).
2007-07-06 06:44:58
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answer #5
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answered by Red_Wings_For_Cup 3
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Linear functions never exceed the first power for the independant variable (x)
2007-07-06 06:43:00
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answer #6
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answered by gfulton57 4
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First of all, it's not a linear function because it contains a squared variable.
Secondly, it is a second degree function as are all parabolas with axis of symmetry parallel to the y-axis.
2007-07-06 06:52:10
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answer #7
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answered by ironduke8159 7
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not a linear function 'cos x^2
linear function => x^1
function 'cos no 2 couples have same 1st element
2007-07-06 06:44:36
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answer #8
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answered by harry m 6
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because its not linear, if you graph it, it will be a parabola, not a line, but its still a relation because it tells you what one variable will be by relating it to the other variable.
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2007-07-06 06:43:41
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answer #9
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answered by Anonymous
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