The top equation doesn't look like a function. It has two outputs for one input.
f(x) = y= ±2x+7
Example: f(2) = ±2(2)+7 = 11 and 3.
Graph it. It won't pass the vertical line test. A vertical line can pass through two points.
The top equation doesn't look like a function. It has two outputs for one input.
f(x) = y= ±2x+7
Example: f(2) = ±2(2)+7 = 11 and 3.
Graph it. It won't pass the vertical line test. A vertical line can pass through two points.
You can substitute in an arbitrary variable like a into the equation as the "input."
This will lead to some "output."
A function can only have one unique output for every input; otherwise, it is not a function.
For example:
f(x) = y= ±2x+7
f(a) = ±2(a)+7 = -2a+7 and 2a+7. Those are two different outputs for that one input; hence, it is not a function.
For example:
f(x) = y= √(5-x)
f(a) = √(5-a) This leads to only one result; hence, it passes the test. This is a function. An input only gives one output.
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Another example:
x^{2} + y^{2} = 1
Is that a function?
Algebraically, we can rewrite it by solving for y.
y^{2} = -x^{2} + 1
±√(y^{2}) = ±√(-x^{2} + 1)
y = ±√(-x^{2} + 1)
This is not a function. The ± tells us that there will be two outputs for a single input.
Again, you can put a in to make sure. y = ±√(-(a)^{2} + 1) = +√(-(a)^{2} + 1) and -√(-(a)^{2} + 1)