Question

[4] Graph of a linear polynomial is straight line. If points (1,-1),(2, 1) and (3/2,0) lie on the graph then find the polynomial and its zero

Answer 1

Answer:

Explanation:

Since (1,-1) lies on the graph, x=1 and y=-1 is a solution to the polynomial. Similarly, x=2 and y=1 is a solution and x=1.5 and y=0 is also a solution.

Since any 2 distinct points have a unique line passing through them, we will only consider the points (1,-1) and (2,1).

A general form of a line is y=mx+b    ...(i)

Here m can be defined as $\frac{y_{1}-y_{2} }{x_{1}-y_{1}}$. Thus, in our case, m = (-1-1)/(1-2) = -2/-1 =2. Thus m = 2.

Substituting m=2 into (i), we get y=2x+b. Since x=-1 and y=1 is a solution of the equation [as (1,-1) lies on the graph], we can substitute these values to get -1=2(1)+b which implies b=-3. Thus, the equation of the polynomial is y=2x-3.

Since it is a linear equation it will have infinitely many zeros.