What if the problem had a negitive number as the leading number such as y = -2x+5?

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I have 3 scenarios for you, given that I'm not fully clear on the intent of your question: 1. y = -2x + 5 is already a function because it's a linear equation. Nothing is raised to a power here, so it's a straight line and the x's and y's already have a 1 to 1 match. 2. If a quadratic equation, which contains x^2 leads with a negative number, it just makes the parabola open downward. 3. The inverse of y = -2x + 5 would be: x = -2y + 5. You only swap the variables, not the signs. You would still want this in y = mx + b form though, so you would just solve for y. x = -2y + 5 Subtract 5 from both sides. x - 5 = -2y Divide by -2. x/-2 - 5/-2 = -2y/-2 -1/2x + 5/2 = y  or, better yet...  y = -1/2x + 5/2. Neither the domain or range would be resricted in any way because this is a linear equation. Hopefully I hit your question with one of these outcomes!