# Provide the linear function if the representing graph passes through the points (2;3) and (-2;-3)?

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The linear function through the given points is a straight line. The equation of the line passing through (x1, y1) and (x2, y2) is: (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1)

Here we have the points (2, 3) and (-2 , -3)

Substituting the values in the equation of the line we get

(y + 3)/( x + 2) = ( 3 + 3)/(2 + 2)

=> (y + 3)/( x + 2) = 3/2

=> 2(y + 3) = 3( x + 2)

=> 2y + 6 = 3x + 6

=> 2y = 3x

=> y = (3/2)x

**The required function is f(x) = (3/2)x**

We'll write the standard form of a linear function f(x):

f(x) = ax + b

We know, from enunciation, that the graph of the function is passing through the given points.

By definition, a point belongs to a curve if the coordinates of the point verify the equation of the curve.

(2;3) is located on the graph of y = ax+b if and only if:

3 = a*(2) + b

2a + b = 3 (1)

(-2;-3) belongs to the graph of y = ax+b if and only if:

-3 = -2a + b

-2a + b = -3 (2)

We'll add (2) to (1)

2a + b -2a + b = 3-3

We'll eliminate and combine like terms:

2b = 0

b = 0

From (1)=>2a+b=3 => 2a = 3

a = 3/2

**The linear function f(x), whose graph is passing through the given points is: f(x) = 3x/2.**