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 3. When completing the square, the constant term being added to both sides should be half of the coefficient of the x term squared, not just 9.
Step 3: h(x)=(x−3)2+2+9 −9
Therefore, the correct vertex is (3, 9).