Complete the square to rewrite y=x^2-6x+16 in vertex form. then state weather the vertex is a minimum and give its coordinates

y = (x - 3)² + 7 is a minimum vertex at (3, 7)

Step-by-step explanation:

the equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

To obtain this form using completing the square

• add/ subtract ( half the coefficient of the x-term )² to x² - 6x

y = x² + 2(- 3)x + 9 - 9 + 16 = (x - 3)² + 7 ← in vertex form

with vertex = (3, 7 )

Since coefficient of x² term > 0 then minimum

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

y > 1/3x

y < = -x + 4

hope this : )