Question

can someone check these for me plz

Answer 1

these this question

Answer 2

please anyone help!!!

Answer 3

For the first one :
y=(x-h)^2 + k
where h,k is the point for your vertex.

In y=3(x+2)^2-3
your h is -2. The reason is because if you plug it in the equation y=(x-h)^2 + k, a negative multiplied to a negative would be + thus getting +2 in y=3(x+2)^2-3 .

Your k is a -3 since the original equation has a +k. A (positive)(negative)=negative, thus getting -3 in place of k in the equation y=3(x+2)^2-3.

So vertex (-2,-3)

Axis of symmetry can be found by x= -(b/2a)
but befoer thatyou have to turn vertex form y=3(x+2)^2-3 to standard form to find b.
In order to do that you have to evaluate that equation to turn it to this form : ax62+bx+c
Step 1: 3(x+2)(x+2)-3 >>Foil x+2
Step 2: 3(x^2+4x+4)-3 >> distribute 3 to what's in parentheses.
Step 3: 3x^2+12x+12-3 >> simplify by adding like terms
= 3x^2+12x+9

Now that you're at stform you can now plug into the equation -(b/2a)
Where a=3 and b =12
-(12/2(3)) = -(12/6)= -(2) = -2

Your axis of symmetry is x=-2

so your answer for the first one would be the second choice vertex = (-2,-3) axis of symm. x=-2

Answer 4

for the first question i posted?

Answer 5

Yep for the vertex problem. I'm still trying to refreshmy head on how to find the max and min. Which follows after the vertex question in the same picture at bam.png

Answer 6

Ah. I don't remember how to find max and min for y=2(x-3)^2 - 4, however I do know that the answer you picked is wrong
2 is a stretch
-3 is a horizontal translation which means you go right 3
-4 is a vertical translation which means you down 4.

That means that -4 is not a domain because your graph is not starting at -4 on the x-axis. -4 would fall into range since you went down 4 times on the y-axis.

That's just to give you insight. I took this class a year or two and I'm forgetting a bunch. I'm sorry!

Answer 7

its ok!