Question

Вырази линейную функцию, график которой параллелен графику линейной функции 4x+2y+3=0 и проходит через точку M(2;3) , через формулу. помогите плез

Answer 1

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Answer 2

Linear Function Parallel to 4x + 2y + 3 = 0 and Passing Through Point M(2, 3)

To express a linear function parallel to the graph of the linear function 4x + 2y + 3 = 0 and passing through the point M(2, 3), we can use the point-slope form of a linear equation, which is given by:

y - y₁ = m(x - x₁)

Where: - (x₁, y₁) is the given point M(2, 3) - m is the slope of the parallel line

First, let's find the slope of the given line 4x + 2y + 3 = 0. The slope-intercept form of a linear equation is y = mx + b, where m is the slope. We can rearrange the given equation to solve for y:

2y = -4x - 3 y = -2x - 3/2

From this, we can see that the slope of the given line is -2.

Now, using the point-slope form, we can express the linear function parallel to the given line and passing through point M(2, 3):

y - 3 = -2(x - 2) y - 3 = -2x + 4 y = -2x + 7

So, the linear function parallel to the graph of 4x + 2y + 3 = 0 and passing through point M(2, 3) is y = -2x + 7.