Question

Функция задана формулой у = 3х + 1. Принадлежат ли графику функции точки А (1; 4) и В (-1; 2)?

Answer 1

$y=3x+1$

$A(1;4)$

$4=3+1$

$4=4=>A$ ∈ $y=3x+1$

$B(-1;2)$

$2=-3+1$

$2$≠$-2$ B∉ $y=3x+1$

Answer 2

To check if points A(1, 4) and B(-1, 2) lie on the graph of the function y = 3x + 1, we need to verify if their coordinates satisfy the equation of the function.

For point A(1, 4): x = 1 y = 4

Let's substitute these values into the function y = 3x + 1: 4 = 3(1) + 1 4 = 3 + 1 4 = 4

Since the equation is true, point A(1, 4) lies on the graph of the function y = 3x + 1.

For point B(-1, 2): x = -1 y = 2

Let's substitute these values into the function y = 3x + 1: 2 = 3(-1) + 1 2 = -3 + 1 2 = -2

The equation is not true for point B(-1, 2), which means point B does not lie on the graph of the function y = 3x + 1.

So, only point A(1, 4) belongs to the graph of the function, while point B(-1, 2) does not.