Question

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.

Answer 1

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).