Find Value of k for Graphs to Enclose Area of 36

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In summary: However, if you meant p(x) = x^2 + 2x + 3 and q(x) = kx + 5x - 7, then you can follow the steps you mentioned to find the value of k for which the graphs enclose an area of exactly 36.
  • #1
Twoacross
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Homework Statement


Determine the value of k for which the graphs of p(x) = x + 2 x + 3 and q(x) = k + 5 x - 7 x
enclose an area of exactly 36. Be sure to use text regions to annotate your work (i.e. explain your reasoning).



Homework Equations





The Attempt at a Solution



So what i tried doing first was finding the points of intersection at a and b to set up my integral and ended up with 3 plus/minus the sqrt(-87) all divided by 16. so I found my a and b values.

Ive attempted in differentiating this question using differential equations methods, and I don't seem to be getting the answer I am looking for.

At this point, I don't know where to go from here, Help would be kindly appreciated!

Thank you!
 
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  • #2
Twoacross said:

Homework Statement


Determine the value of k for which the graphs of p(x) = x + 2 x + 3 and q(x) = k + 5 x - 7 x
enclose an area of exactly 36. Be sure to use text regions to annotate your work (i.e. explain your reasoning).



Homework Equations





The Attempt at a Solution



So what i tried doing first was finding the points of intersection at a and b to set up my integral and ended up with 3 plus/minus the sqrt(-87) all divided by 16. so I found my a and b values.

Ive attempted in differentiating this question using differential equations methods, and I don't seem to be getting the answer I am looking for.

At this point, I don't know where to go from here, Help would be kindly appreciated!

Thank you!

Your first step would be to state the problem correctly. Maybe you need to use the X2 button. You have given two strangely written equations of straight lines which, by themselves, can't enclose an area.
 

1. What is the formula for finding the value of k for graphs to enclose an area of 36?

The formula for finding the value of k for graphs to enclose an area of 36 is k = √(36/A), where A is the area of the graph. This formula can be derived from the formula for finding the area of a rectangle, A = l × w, where l is the length and w is the width. In this case, we are looking for a square graph with an area of 36, so the length and width are both equal to √36 = 6. Therefore, the value of k is 6.

2. How is the value of k related to the area of the graph?

The value of k is directly proportional to the area of the graph. This means that as the area of the graph increases, the value of k also increases. In other words, a larger area requires a larger value of k to enclose it.

3. Can the value of k be negative?

No, the value of k cannot be negative. It represents the length of the sides of the square graph, and length cannot be negative. If the area of the graph is negative, it means that the graph is flipped or rotated, but the value of k will always be positive.

4. How does the shape of the graph affect the value of k?

The shape of the graph does not affect the value of k as long as the area is 36. This is because the formula for finding the value of k only takes into account the area of the graph, not its shape. As long as the area is 36, the value of k will always be 6.

5. Can the value of k be a decimal or a fraction?

Yes, the value of k can be a decimal or a fraction. As long as the result of the formula k = √(36/A) is a positive number, it is a valid value for k. This means that the value of k can be a whole number, a decimal, or a fraction, depending on the area of the graph.

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