One way to test if a graph is a function or not is by doing a?

Mathematics

2 answers:

KATRIN_1 [288]2 years ago

7 0

One way to test if a graph is a function or not is by doing a vertical line test.

The vertical line test is tested by testing if any vertical line is drawn would intersect the curve more than once. If there is any such line, the graph does not represent a function, if there isn't any such line, the graph is a function.

Hope that helps!

fenix001 [56]2 years ago

4 0

Answer:

Vertical line test. (literally what it say)

Step-by-step explanation:

Given a graph, use the vertical line test to determine if the graph represents a function. Inspect the graph to see if any vertical line drawn would intersect the curve more than once.

Therefore you would have x value being the same for several points.

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You have a puppy named Fido and 60 feet of fencing. You want to make him the largest play area possible. What are the dimensions

LUCKY_DIMON [66]

Answer:

15 feet x 15 feet

Step-by-step explanation:

I'm worried about Fido, so I'll use three different approaches to maximizing his play space: Graphing, Chart, and First Derivative. This is perhaps more than Fido wants, but it hopefully offers ideas on how these types of problems can be answered.

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Fido is limited to 60 feet of fence, which will form the perimeter. The area of a rectangle is (length)*(width). Let's set length to L and width to W. I'll use A for area.

Therefore A = L*W

The rectangle's perimeter will be 2L + 2W, and this is equal to 60 feet:

2L + 2W = 60

Let's rearrange this equation to isolate either W or L. I chose L:

L + W = 30

L = 30 - W

Now let's use this definition of L in the Area equation:

A = L*W

A = (30-W)*W

A = 30W - W^2

Graphing

We can graph this function (A = 30W - W^2) to find the maximum area. See the attached image. We find a maximum area of 225 ft^2 when the width is 15 feet. 15 feet would mean that Fido's roaming pen will be in the shape of a square; 15 feet for each side equals 60 feet total fence.

Chart

One can also find a maximum by calculating a variety of conditions set by L = 30 - W. I randomly picked lengths from 2 to 24 feet and calculated the resulting width and then the resultant area for each condition. See attached image. Although I did not pick 15 feet, we can see that the area increases until the length transitions from 14 to 16 feet. It reverses direction at that point. A closer inspection around those values will show that 15 feet is, in fact, the optimum value to produce a pen with the most area.

First Derivative

One could also take the first derivative of the equation A = 30W - W^2 and set it equal to zero. The fist derivative tells us the slope of the line at any point (for a value of W in this example). The slope is zero only when the curve hits a maximum.

A = 30W - W^2

A' = 30 -2W

0 = 30 - 2W

2W = 30

W = 15 feet

The slope is zero when the width is 15 feet. 15 feet is the optimum for creating a pen with the most area possible with 60 feet of fence.

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All three approaches yield the same result. Graphing is fun when you have access to DESMOS. Charts are fun if you like Excel. Derivatives can also be fun, mostly because I always ponder "Who had the audacity to think this might work?" (Didn't they have something more important to do? Raid a village, etc.?).

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Arf