SOLUTION: Find an equation for the lower half of the parabola, {{{(y+1)^2=x+3)}}}. Can someone help walk me through the steps of figuring this out please?

Algebra ->  Trigonometry-basics -> SOLUTION: Find an equation for the lower half of the parabola, {{{(y+1)^2=x+3)}}}. Can someone help walk me through the steps of figuring this out please?      Log On


   

Question 895998: Find an equation for the lower half of the parabola, %28y%2B1%29%5E2=x%2B3%29. Can someone help walk me through the steps of figuring this out please?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
equation is (y+1)^2 = x+3

take the square root of both sides of this equation to get:

y+1 = plus or minus sqrt(x+3)

subtract 1 from both sides of this equation to get:

y = -1 plus or minus sqrt(x+3)

the upper half of the parabola is y = -1 plus sqrt(x+3)

the lower half of the parabola is y = -1 minus sqrt(x+3)

here's the graph of the upper half which is y = -1 + sqrt(x+3).

graph%28600%2C600%2C-10%2C10%2C-10%2C10%2C-1%2Bsqrt%28x%2B3%29%29

here's the graph of the lower half which is y = -1 - sqrt(x+3).

graph%28600%2C600%2C-10%2C10%2C-10%2C10%2C-1-sqrt%28x%2B3%29%29

here's the graph of both halves which is y = -1 + sqrt(x+3) and the graph of y = -1 - sqrt(x+3).

graph%28600%2C600%2C-10%2C10%2C-10%2C10%2C-1%2Bsqrt%28x%2B3%29%2C-1-sqrt%28x%2B3%29%29