Recent content by grain1

Thank you again Lanedance, you are being very helpful. I thought the min was at the vertex... i.e. -9? g(x) = 1/4 (x - 1)^2 - 9 (1 ≤ x ≤ 7) g^{-1}(y) = 1/4 (y - 1)^2 - 9 x = 1/4 (y - 1)^2 - 9 4x + 36 = (y - 1)^2 √(4x + 36) = y - 1 +2x + 6 + 1...

Thanks Lanedance... -8 3/4 is the y - intercept, does this mean my image will be (-8 3/4, ∞)? SammyS ... yes i meant to type 1≤x≤7. HallsofIvy... i thought the inverse was found by switching the x and y in the oringinal formula i.e y = 1/4 (x-1)^2 - 9 becomes x = 1/4 (y-1)^2 - 9 and...

[b]1. f(x) = 1/4 (x - 1)^2 - 9. What is the image set of the function f? Express in interval notation. g(x) = 1/4 (x - 1)^2 - 9. (1 ≤ x ≥ 7). Specify the domain and image set of the inverse function g^-1, and find it's rule. I really don't understand what happens in these questions. I...