# For the function y = x^2  - 2x - 5 determine:  a) it's vertex. [use (,) with coordinates] b) the x intercepts in simplified form.`` Hi, Kristen, To solve this, for the vertex, there is a formula we can use: x = -b/(2a) This gives the x coordinate of the vertex. a and b come from the general formula y = ax^2 + bx + c. Comparing your function to that formula, a = 1 and b = -2 (c = -5). So, plugging in: x = -(-2)/(2*1) = 1 So, x = 1. To find the y, we can always plug the value of x back into the original equation. So, plugging in 1 for x: y = 1^2 - 2*1 - 5 = 1 - 2 - 5 = -6 Therefore, the vertex is at (1,-6) For the x intercepts, all x intercepts have y = 0. Therefore, we would actually be solving: 0 = x^2 - 2x - 5 It doesn't factor, so we have to solve using the quadratic formula or completing the square. To complete the square, we first make sure if a = 1. It is, so then, we move c to the other side. Here, we add 5: 5 = x^2 - 2x Then, we take half of b, -2/2 = -1, then square that result, (-1)^2 = 1. We then add that result to each side: 5+1 = x^2 - 2x + 1 The right side factors 6 = (x-1)^2 Taking the square root of each side then adding 1 to each side: 1 +- sqrt 6 = x, or x = 1 +- sqrt 6 = approx 3.45 and -1.45 So, the x intercepts would be (3.45,0) and (-1.45,0) Good luck, Kristen. I hope this helps. Till Then, Steve
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