Post a New Question

Algebra

posted by .

I have an assignment to write an equation to point-slope form, then to standard form, and then to slope-intercept form. I have used the points (-2,-1) or (1,2), but I used (-2,-1).

Now, I have the point-slope form of y + 1= 1(x + 2) and the standard form of y – 1x = -1. When I convert into slope-intercept form, I am getting the slope as -1, which is the problem. Could someone please tell me where I am going wrong?

  • Algebra -

    Actually, the slope is not -1, but the y-intercept comes up as -1.

  • Algebra -

    your slope is correct as +1
    your equation of
    y+1 = 1(x+2) is also correct

    then y+1 = x+2
    y = x+1 ---> slope-yintercept form

    y-x = 1
    x - y = -1 ----> standard form ( I multiplied each term by -1)

  • Algebra -

    So, how did you multiply the terms by -1?

  • Algebra -

    And how do you rewrite x - y = -1 into slope-intercept form?

  • Algebra -

    in standard form we usually lead with the x term with a positive coefficient, so I wanted a +x as my first term, it was -x .

    to change a standard form to slope- yintercept form usually takes 2 steps:
    1. move all terms except the y term to the right
    e.g.
    4x - 5y = 8 ---≥ -5y = -4x + 8
    2. divide each term by the coefficient of the y term
    -5y = -4x + 8 ----> y = (-4/-5)x + 8/-5
    y = (4/5)x - 8/5 , just an example, has nothing to do with your question

    apply those 2 steps to your question.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question