What is the equation of the line that passes through (4, -1) and (-2, 3)?

Mathematics

1 answer:

Mrac [35]3 years ago

7 0

Answer:

2x + 3y - 5 = 0

Step-by-step explanation:

(4 , -1) & (-2,3)

$Slope =\frac{y_{2}-y_{1}}{x_{2}-x_1}}\\\\=\frac{3-[-1]}{-2-4}\\\\=\frac{3+1}{-6}\\\\=\frac{4}{-6}\\\\=\frac{-2}{3}$

m = -2/3; (4 , -1)

y - y1 = m(x - x1)

$y - [-1]= \frac{-2}{3}(x-4)\\\\y + 1 =\frac{-2}{3}x -4*\frac{-2}{3}\\\\y + 1 =\frac{-2}{3}x+\frac{8}{3}$

Multiply the equation by 3

3y + 3 = -2x + 8

3y = -2x + 8 - 3

3y = -2x + 5

2x + 3y - 5 = 0

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100 points (algebra one) describe and correct the error in comparing the graphs

Pepsi [2]

Answer:

The error made is that the graph of $y=x^{2} +3$ is a vertical translation up by 3, instead of down.

Step-by-step explanation:

Exponential functions, when they are graphed, use the function $f(x)=a(x-h)^{2} +k$ .

In the equation, k expresses how much the graph shifts vertically. In this case, the function shifts vertically up by 3 from the parent function, $y=x^{2}$ . So, the translation is actually 3 units up compared to the graph of $y=x^{2}$ .