Question

Найдите линейную функцию, график которой параллелен графику линейной функции y=--3x+1 и проходит через точку А(--2;10). Постройте график полученной функции. Пожалуйста!! Очень срочно!!

Answer 1

У паралллельных прямых равные угловые коэффициенты, поэтому к=-3.

$k=-3,\; A(-2,10)\\\\y-y_0=k(x-x_0)\\\\y-10=-3(x+2)\\\\y=-3x+4$

Answer 2

Finding the Linear Function

To find a linear function that is parallel to the graph of the given linear function y = -3x + 1 and passes through the point A(-2, 10), we can use the fact that parallel lines have the same slope.

The slope of the given linear function is -3. Therefore, the slope of the parallel line will also be -3.

Using the point-slope form of a linear equation, we can write the equation of the parallel line as:

y - y1 = m(x - x1)

where (x1, y1) is the given point A(-2, 10) and m is the slope of the line.

Substituting the values, we have:

y - 10 = -3(x - (-2))

Simplifying the equation, we get:

y - 10 = -3(x + 2)

Now, let's expand and simplify the equation further:

y - 10 = -3x - 6

Adding 10 to both sides of the equation, we get:

y = -3x + 4

Therefore, the linear function that is parallel to the graph of y = -3x + 1 and passes through the point A(-2, 10) is y = -3x + 4.

Graphing the Function

To graph the function y = -3x + 4, we can plot a few points and draw a straight line through them.

Let's choose some x-values and calculate the corresponding y-values:

When x = -3, y = -3(-3) + 4 = 13 When x = -2, y = -3(-2) + 4 = 10 When x = -1, y = -3(-1) + 4 = 7 When x = 0, y = -3(0) + 4 = 4 When x = 1, y = -3(1) + 4 = 1

Plotting these points (-3, 13), (-2, 10), (-1, 7), (0, 4), and (1, 1) on a graph and connecting them with a straight line, we get the graph of the function y = -3x + 4.


Please note that the graph is a straight line with a slope of -3 and a y-intercept of 4.