The vertex of this parabola is at (-3, 6). which of the following could be its equation?
A. y=-3(x-3)^2+6
B. y=-3(x+3)^2+6
C. y=-3(x-3)^2-6
D. y=-3(x+3)^2-6

Option B -  y=-3(x+3)^2+6

Step-by-step explanation:

Step 1:

The equation of any parabola in the standard form with vertex at the origin is y^2 = 4ax  . When it opens upward, the equation would be x^2 = 4ay

When the vertex is shifted to a point other than the origin, the equation would be

(x-h)^2 = 4a(y-k)

Step 2 :

Given that the vertex of the parabola is at  (-3, 6). We will have the x and y co-ordinate shifted with this value.

So the equation of the parabola will have x and y shifted to x-(-3)) and y-6

=> x+3 and y-6

Step 3:

Based on the above 2, we can see that option B  y=-3(x+3)^2+6 could be the equation of the parabola.

∠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.

idk wbu

step-by-step explanation: