# For a linear function f, f(-1)=3 and f(2)=4 find an equation for fplease answer soon, I need help badly

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Another way to answer this question is that you know any linear function can always be written y = mx + b, where m is the slope.

Find m first, (4 - 3)/(2 - -1)=1/3.

Then plug in one point to solve for b, 4 = (1/3)(2) + b, solve for b and b = 10/3.

Replace m and b with their values, and y = (1/3)x + (10/3).

Here is another example where we find the equation of a line from two points:

The general form of a linear function is

f(x)=ax + b

So, f(-1)=3 means that we hhave to substitute, in the general form, the unknown x with the value -1.

-1xa + b = 3

We'll do the same thing with the following f(2)=4

2a +b = 4

Now, in order to find the function f, we have to determine the unknown coefficients a and b.

SO, we'll subtract from the last relation, 2a +b = 4, the anterior one ,-a + b= 3.

2a +b - (-a) +b = 4-3

3a = 1

a = 1/3

With the known value of a, we are going in any of the both anterior relation, substitute a and find out the value of b unknown.

We choose the relation -a + b =3, beccause it's more simple to calculate b, after substituting a value.

-1/3 + b = 3

b = 3 + 1/3

The common denominator is 3, so we have to multiply the first term 3, with the common denominator 3

b = ( 9 +1)/3

b = 10/3

Now we could write down the linear function f.

f(x) = (1/3)x + 10/3

General equation of linear function f is:

f(x)=mx+c

Since f(-1)=3, so we have to subsitute x=-1

-m+c=3 -eqn 1

Also, f(2)=4, so we have to sub. x=2

2m+c=4 -eqn 2

Subtract eqn 1 from eqn 2

2m-(-m)+c-c=4-3

3m=1

m=1/3

subsitute m=1/3 into eqn 2

2(1/3)+c=4

2/3+c=4

c=4-2/3

=12/3-2/3

= 10/3 or 3 1/3

The equation of f(x): 1/3 x+10/3 (mx+c)