How do you find the vertex for # y=x^2-x-2#?

Answer 1

Genesis Allen

Vertex is at # (1/2 , -2 1/4) #

# y= x^2 -x -2 or y= x^2 -x +(1/2)^2 -1/4 -2 # or

# y= (x-1/2)^2 -9/4 # , Comparing with vertex form of equation

#y= a (x-h)^2+k ; (h,k)# being vertex , we find here

#h=1/2 , k = -9/4# . So vertex is at # (1/2 , -2 1/4) #

graph{x^2-x-2 [-10, 10, -5, 5]} [Ans]

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Answer 2

Eva Adamson

To find the vertex of a quadratic equation in the form y = ax^2 + bx + c, you can use the formula x = -b / (2a) to find the x-coordinate of the vertex. Then substitute this value into the equation to find the corresponding y-coordinate. For the given equation y = x^2 - x - 2, a = 1, b = -1. Using x = -b / (2a), we find x = -(-1) / (2 * 1) = 1/2. Substitute x = 1/2 into the equation to find y: y = (1/2)^2 - (1/2) - 2 = -9/4. So, the vertex is (1/2, -9/4).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.