How do you find out your vocal range?

How do you find the domain and range of a original and inverse quadratic function?

• My teacher has never used the terms vertex none of those things i have found my equation y= (x-3)^2 and i solved it simultaneously to get x^2+6-x Now how do i find the domain and range from this? Can you actually find the domain and range from a formula? I was going to use the above equation in a quadratic equation and find the two points, which i would assume would be my domain and range? Or would i have to do something else? IM SO CONFUSED!!! I have no idea what to do, my teacher has explained it heaps of times and i still dont understand i am currently sick and have spent all day looking up the answer on the web and math tutoring sites, so could SOMONE please please help me!!! THANKU :)

• Answer:

  y = (x - 3)² y = (x - 3)(x - 3) y = x² - 3x - 3x + 9 y = x² - 6x + 9 [or you could use the rule (a - b)² = a² - 2ab + b² and write the expansion x² - 6x + 9 straight down] now basically the domain are the values of x for which the function is defined y = x² - 6x + 9 is a parabola that opens upwards and there are no values of x for which the function is not defined ... [b/c there is no denominator with x in it] so the domain of y = x² - 6x + 9 is all values of x ... so if they want the answer in interval notation the domain is (-∞, +∞) the range is all the possible values of y B/c the parabola opens upwards it has a minimum turning point ... (at the vertex) The minimum y-value is the y-coordinate of the vertex the x-coordinate of the vertex is given by x = -b/2a = 6/2 = 3 and the y-coordinate of the vertex is y = 3² - 6 * 3 + 9 y = 0 so the parabola has it's lowest point at (3, 0) ... so it doesn't go below the x-axis and the range is y ≥ 0 ... so in interval notation the range is [0, +∞)