Cole rewrote a quadratic function in vertex form.
h(x) = x²-6x+7
Step 1: h (x) = (x²-6x+)+7
Step 2: h (x) = (x²-6x+9)+7-9
Step 3: h (x) = (x-3)²+2
Cole said that the vertex is (3, 2). Is Cole correct? If not, identify the step in which Cole made the mistake and correct his work.
Cole is incorrect. The mistake is in step 2 where Cole subtracted 9 from 7 instead of adding 9 to complete the square.
The corrected version should be:
Step 1: h(x) = (x² - 6x) + 7
Step 2: h(x) = (x² - 6x + 9) + 7 - 9
Step 3: h(x) = (x - 3)² - 2
Therefore, the correct vertex of the quadratic function h(x) = x² - 6x + 7 is (3, -2), not (3, 2) as Cole stated.