Function equation, Determine

A linear function is uniquely determined by two of its value pairs or by two points of its graph. If one of the given pairs of values is the pair (0; 0), i.e. if the graph of the function passes through the origin of the coordinates, then finding the equation is particularly easy - pay for homework . A linear function is uniquely determined by two of its value pairs or by two points of its graph.

If one of the given pairs of values is the pair (0; 0), i.e. if the graph of the function passes through the origin of the coordinates - algerbra homework help , then finding the equation is particularly easy.

1st way:

Since the graph intersects the y-axis at the point 0, the parameter n in the equation y=f(x)=mx+n has the value zero. Therefore y=mx and m=yx are valid - math homework help . Using the coordinates (x; y) of a second given point of the graph, m can thus be calculated, since these coordinates must satisfy the equation y=f(x)=mx+n.

Examples:

The graph of a linear function runs

a) through the coordinate origin and the point P1(2; 4);
b) through the origin and the point P2(3; -4).

In each case the equation of the function is to be set up.

a) From m=yx follows m=42=2 . Thus applies:
y=f(x)=2x

b) From m=yx follows m=-43=-43 . Thus holds:
y=f(x)=-43x

2nd way:

We enter a rise triangle in the graph of the function given by the two points and calculate from its cathetus lengths (at least approximately) the ratio m=yx. Referring to the examples above, we get from figure 1

a) m=42=-6-3=2, thus y=f(x)=2x

b) m=4-3=-43=-43, thus y=f(x)=-43x

The graph of a linear function uniquely determined by two points does not pass through the coordinate origin.

Example:

Let P1(2; 5)and P2(-2; -1) be two points of the graph of the linear function y = f(x) = mx + n. The equation of the function is to be set up.

1st way:

From Figure 2, we can see that the straight line intersects the y-axis at the point 2. So n = 2 and therefore y=f(x)=mx+2. The value of m can again be found using the cathetus lengths of a rise triangle. In our case applies: m=3:2=32

Another consideration is also possible: the coordinates of P1(and also of P2) must satisfy the equation of f. So 5=m⋅2+2 is valid, from which we also get m=32.
The equation of the function f is thus y = f(x) = x + 2.

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