y = -2x² + 8x + 3 Factor the coefficient of the x² term out of the x-terms on the right [Do not factor out the x, but only the numerical coefficient] y = -2(x² + 4x) + 3 Multiply the coefficient of x by 1/2, then square what you get. 4(1/2) = 2, then 2² = 4 Add and then subtract that number inside the parentheses: [This just amounts to adding 0 which does not change the value] y = -2(x² + 4x + 4 - 4) + 3 Change the parentheses to brackets so you can insert parentheses around the first three terms: y = -2[(x² + 4x + 4) - 4] + 3 Factor the trinomial in the parentheses which should turn out to be the square of a binomial: y = -2[(x + 2)(x + 2) - 4] + 3 y = -2[(x + 2)² - 4] + 3 Remove the brackets by multiplying the -2 by each of the terms inside the bracket, keeping the (x + 2)² intact: y = -2(x + 2)² + 8 + 3 y = -2(x + 2)² + 11 That's it. Compare to y = a(x - h)² + k and you will see that the vertex is (-2,11) ------------------------------------------- y = 3x² - 6x + 5 Factor the coefficient of the x² term out of the x-terms on the right [Do not factor out the x, but only the numerical coefficient] y = 3(x² - 2x) + 5 Multiply the coefficient of x by 1/2, then square what you get. -2(1/2) = -1, then (-1)² = 1 Add and then subtract that number inside the parentheses: [This just amounts to adding 0 which does not change the value] y = 3(x² - 2x + 1 - 1) + 5 Change the parentheses to brackets so you can insert parentheses around the first three terms: y = 3[(x² - 2x + 1) - 1] + 5 Factor the trinomial in the parentheses which should turn out to be the square of a binomial: y = 3[(x - 1)(x - 1) - 1] + 5 y = 3[(x - 1)² - 1] + 5 Remove the brackets by multiplying the 3 by each of the terms inside the bracket, keeping the (x + 2)² intact: y = 3(x - 1)² - 3 + 5 y = 3(x - 1)² + 2 That's it. Compare to y = a(x - h)² + k and you will see that the vertex is (1,2) Edwin