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Ivan
3 years ago
7

When given a graph, the vertical line test can be used to determine functionality. Describe the vertical line test and explain t

he reasons why a graph would, or would not, represent a function.
Mathematics
2 answers:
yulyashka [42]3 years ago
5 0

Answer:

When given a graph, the vertical line test can be used to determine functionality. The vertical line test would represent a function if no more than one point of the graph of a relation, this shows that only one output value for each input value. If this doesn't apply or more than one point is on the line then it would not represent a function.

Step-by-step explanation:

Draw vertical lines to intersect on a graph.

The relation is a function if there’s only one point of intersection on a graph.

The relation is not a function if there’s more than one point of intersection on a graph.

The vertical line test is used to determine if each input has exactly one output.

diamong [38]3 years ago
3 0

Answer:

The vertical line test is a way for you to see if a graph represents a function. It allows you to identify if any x values have more than one y value.

A graph would be a function if every input (x) has exactly one output (y). A graph would not be a function if an input (x) has more than one output (y).

Step-by-step explanation:

In a function, every input within the domain of the function must have exactly one output. If the graph has an input that has more than one output, then it is not a function. The vertical line test is what allows you to see if a graph is a function or not.

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Leto [7]

The value of f(a)=4-2a+6a^{2}, f(a+h) is 6a^{2} +6h^{2} -2a-2h+12ah , [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6x^{2}.

Given a function f(x)=4-2x+6x^{2}.

We are told to find out the value of f(a), f(a+h) and [f(a+h)-f(a)]/hwhere h≠0.

Function is like a relationship between two or more variables expressed in equal to form.The value which we entered in the function is known as domain and the value which we get after entering the values is known as codomain or range.

f(a)=4-2a+6a^{2} (By just putting x=a).

f(a+h)==4-2(a+h)+6(a+h)^{2}

=4-2a-2h+6(a^{2} +h^{2} +2ah)

=4-2a-2h+6a^{2} +6h^{2} +12ah

=6a^{2} +6h^{2}-2a-2h+12ah

[f(a+h)-f(a)]/h=[6a^{2} +6h^{2}-2a-2h+12ah-(4-2a+6a^{2} )]/h

=(6a^{2} +6h^{2} -2a-2h+12ah)/h

=(6h^{2} -2h+12ah)/h

=6h+12a-2.

Hence the value of function f(a)=4-2a+6a^{2}, f(a+h) is 6a^{2} +6h^{2} -2a-2h+12ah , [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6x^{2}.

Learn more about function at brainly.com/question/10439235

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