PLEASE HELP 25 POINTS Write the equation for the following relation. Include all of your work in your final answer. Submit your solution. R = {(x, y): (2, 3), (4, 4), (6, 5), (8, 6), . . .}

Respuesta :

R = {(x, y): (4, 5), (8, 7), (12, 9), (16, 11), . . .}

Notice that y increases by 2 when x increases by 4

slope = 2/4 = 1/2

Notice that if you decrease x by 4 y will decrease by 2

So, y-intercept = (0,3)

Equation: y = (1/2)x+3

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R = {(x, y): (2, 3), (4, 4), (6, 5), (8, 6), . . .}

slope = 1/2

y-int: (0,1)

Equation: y = (1/2)x + 1

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Q = {(x, y): (2, 8), (3, 27), (4, 64), (5, 125), . . .}

Not a linear relation because y does not increase consistently

as x increases.

Equation: y = x^3

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P = {(x, y): (1, 0), (2, 4), (3, 8), (4, 12), . . .}

slope = 4

y-int: -4

Equation: y = 4x-4

Answer:  The required equation is [tex]y=\dfrac{1}{2}x+2.[/tex]

Step-by-step explanation:  We are given to write the equation for the following relation :

R = {(x, y) : (2, 3), (4, 4), (6, 5), (8, 6), . . .}

Since each value of x is associated to one and only one value of y, so there will be a linear relation between x and y.

That is, the relation will make a straight line when drawn on the co-ordinates grid.

The line passes through the points (2, 3) and (6, 5), so the slope of the line will be

[tex]m=\dfrac{5-3}{6-2}=\dfrac{2}{4}=\dfrac{1}{2}.[/tex]

Also, the line passes through the point (2, 3), so its equation is given by

[tex]y-3=\dfrac{1}{2}(x-2)\\\\\\\Rightarrow y=\dfrac{1}{2}x-1+3\\\\\Rightarrow y=\dfrac{1}{2}x+2.[/tex]

Thus, the required equation is [tex]y=\dfrac{1}{2}x+2.[/tex]

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