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

you didn't change any of the steps they are the exact same as I put them

Answer 2

Cole made a mistake in Step 2 when he added 9 inside the parentheses. Instead, he should have added (9) outside the parentheses after the +7.

Corrected steps:

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 is (3, 2). Cole's answer is correct.

Answer 3

You are correct, I apologize for the oversight. Let's correct Cole's steps:

Step 1: h(x)= (x2−6x+_) +7

Step 2: h(x)= (x2−6x+9)+7 -9

Step 3: h(x)= (x−3 )2+7 -9

Step 3 should be corrected as:

h(x)= (x−3 )2 - 2

Therefore, the correct vertex is (3, -2), not (3, 2). Cole's answer was incorrect, and the mistake was made in Step 3 where 7 should have been subtracted from 9 instead of 9 being subtracted from 7.