What is the equation of the line, in slope-intercept form, that passes through (3, -1) and (-1,5)?

2 answers:

8 0

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{5 - (-1)}{-1 - 3} = \frac{5 + 1}{-4} = \frac{6}{-4} = \frac{3}{-2} = -1\frac{1}{2}$
y - y₁ = m(x - x₁)
y - (-1) = -1¹/₂(x - 3)
y + 1 = -1¹/₂(x) + 1¹/₂(3)
y + 1 = -1¹/₂x + 4¹/₂
      - 1 - 1
   y = -1¹/₂x + 3¹/₂

4x + y - 2 = 0
      + 2 + 2
       4x + y = 2
4x - 4x + y = -4x + 2
   y = -4x + 2
   y - y₁ = m(x - x₁)
      y - (-3) = -4(x - 4)
       y + 3 = -4(x) + 4(4)
   y + 3 = -4x + 16
       - 3 - 3
   y = -4x + 13

2x - 3y - 5 = 0
       + 5 + 5
2x - 2x - 3y = -2x + 5
   -3y = -2x + 5
      -3    -3
      y = ²/₃x - 1²/₃
   y - (-3) = -1¹/₂(x - 4)
      y + 3 = -1¹/₂(x) + 1¹/₂(4)
      y + 3 = -1¹/₂x + 6
      - 3     - 3
      y = -1¹/₂x + 3

Kaylis [27]3 years ago

4 0

its b because the others are incorrect

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For what values of $x$ is $$\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0?$$ note: be thorough and explain why all points in your answe

Rus_ich [418]

To solve the inequality $\frac{x^2+x+3}{2x^2+x-6}\ge \:0$

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The numerator $x^2+x+3$ is not factorizable.

so factor the denominator $2x^2+x-6$ :

$2x^2+x-6=\left(2x^2-3x\right)+\left(4x-6\right)=x\left(2x-3\right)+2\left(2x-3\right)\\ \mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-3x\mathrm{:\quad }x\left(2x-3\right)\\ \mathrm{Factor\:out\:}2\mathrm{\:from\:}4x-6\mathrm{:\quad }2\left(2x-3\right)\\$