Answer to Question #251056 in Algebra for Iska

Question

J(t) = 1/t²-2t+1

Answer 1

Question

Construct a table of values of the following functions using the interval of -5 to 5.

$j(t)=\frac{1}{t^2-2t+1}$

solution

The points lies between the interval (−5,5)  are −4,−3,−2,−1,0,1,2,3,4

Substitute t=−4 in the function $j(t)=\frac{1}{t^2-2t+1}$

$j(-4)=\frac{1}{(-4)^2-2(-4)+1}\\=\frac{1}{16+8+1}\\=\frac{1}{25}$

Substitute t=−3 in the function $j(t)=\frac{1}{t^2-2t+1}$

$j(-3)=\frac{1}{(-3)^2-2(-3)+1}\\=\frac{1}{9+6+1}\\=\frac{1}{16}$

Substitute t=−2 in the function $j(t)=\frac{1}{t^2-2t+1}$

$j(-2)=\frac{1}{(-2)^2-2(-2)+1}\\=\frac{1}{4+4+1}\\=\frac{1}{9}$

Substitute t=−1 in the function $j(t)=\frac{1}{t^2-2t+1}$

$j(-1)=\frac{1}{(-1)^2-2(-1)+1}\\=\frac{1}{1+2+1}\\=\frac{1}{4}$

Similarly, find the other values of j(0),j(1),j(2),j(3),j(4)

j0,j1,j2,j3,j4.

Make a table of values of the given functions using the interval of -5 to 5.