vertex form of 3x2-x-10

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It looks like you are trying to rewrite the following quadratic equation in vertex form.y = 3x^2 - x - 10First...pull out a factor of 3 from each term on the right.y = 3(x^2 - 1/3x - 10/3)Now, complete the square for the expression in parentheses.y = 3(x^2 - 1/3x + 1/36 - 10/3 - 1/36)notice that we just add and subtract the square of half of the coefficient of the x-term. The first three terms are now a perfect square, namely (x - 1/6)^2y = 3((x - 1/6)^2 - 10/3 - 1/36)Combine the like terms in the parentheses.y = 3((x - 1/6)^2 - 121/36)Finally, distribute the factor of 3 through the quantity in parentheses.y = 3(x - 1/6)^2 - 121/12The vertex is at (1/6, -121/12)That should do it for you.