Cole rewrote the quadratic function h(x) = x^2 - 6x

Cole rewrote a quadratic function in vertex form. h(x)=x^2 -6x+7Cole said that the vertex is (3, 2). Is Cole correct? If not,
1. 1 answer
2. 48 views
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
1. 1 answer
2. 62 views
Cole rewrote a quadratic function in vertex form.h(x)= x^2−6x+7 Step 1: h(x)= (x^2−6x+ )+7 Step 2: h(x)=(x^2−6x+ 9 )+7
1. 3 answers
2. 42 views
Cole rewrote a quadratic function in vertex form.h(x)= x^2−6x+7 Step 1: h(x)= (x^2−6x+ )+7 Step 2: h(x)=(x^2−6x+ 9 )+7
1. 1 answer
2. 41 views
Cole rewrote a quadratic function in vertex form.h(x)= x^2−6x+7 Step 1: h(x)= (x^2−6x+ )+7 Step 2: h(x)=(x^2−6x+ 9 )+7
1. 5 answers
2. 38 views
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
1. 3 answers
2. 51 views
Cole rewrote a quadratic function in vertex form.h(x)= x2−6x+7 Step 1: h(x)= (x2−6x+ 9 )+7 Step 2: h(x)=(x2−6x+ 9 )+7 −9
1. 1 answer
2. 132 views
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
1. 1 answer
2. 59 views
Cole rewrote the quadratic function h(x) = x^2 - 6x +7 into vertex form, Cole said the vertex is (3.2) is he correct? Show your
1. 3 answers
2. 45 views
Cole rewrote a quadratic function in vertex form. h(x)=x^2 -6 +7
1. 1 answer
2. 25 views