Question

Найдите линейную функцию, график которой параллелен графику линейной функции у=-3х+1 и проходит через точку А (-2;10). ПОМОГИТЕ, ПОЖАЛУЙСТА!

Answer 1

Y = -3x +b. подставляем вместо х и у координаты точки А, получится 
10 = -3*(-2) + b, следовательно b = 4. Поэтому уравнение прямой, которая параллельна данной и проходит через точку а равно у= -3х + 4

Answer 2

Finding the Linear Function Parallel to \(y = -3x + 1\) and Passing Through Point A(-2, 10)

To find the linear function parallel to \(y = -3x + 1\) and passing through point A(-2, 10), we can use the fact that parallel lines have the same slope. The slope of the given line \(y = -3x + 1\) is -3. We can use this slope and the given point A(-2, 10) to find the equation of the parallel line.

The general form of a linear function is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.

Steps to Find the Linear Function:

1. Find the Slope of the Parallel Line: - Since the parallel line has the same slope as \(y = -3x + 1\), the slope of the parallel line is also -3.

2. Use the Point A(-2, 10) to Find the y-Intercept: - We can use the point-slope form of a linear equation to find the y-intercept. The point-slope form is given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the given point and \(m\) is the slope.

3. Write the Equation of the Parallel Line: - Once we have the slope and the y-intercept, we can write the equation of the parallel line in the form \(y = mx + b\).

Solution:

The equation of the linear function parallel to \(y = -3x + 1\) and passing through point A(-2, 10) is:

\[y = -3x + 4\]

This linear function has a slope of -3 (same as the given line) and passes through the point A(-2, 10).

Let me know if there's anything else you'd like to discuss or any other questions you have!