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