1. ## hi im new in 8th grade..need some homework help

write the equation of the axis of symetry and find the coordinates of the vertex of the graph of each equation
12.y=-3x^2 +4

y=3(x+1)^2 -20

y=-3x^2+4

y=3x^2+6x-17

y=-x^2+2x

2. I do not answer homeworks that have more than two problems. You have 5 problems here, so I should not answer your question here. But then, maybe I can show solutions for two problems only and you do other three.

There are many ways to solve the problems, which are all vertical parabolas. My favorite way is by transforming the equation into a standard form of a vertical parabola, but if you have not studied that yet, then you won't understand the solution.

Another way is by using the formula x = -b/(2a) for the x-coordinate of the vertex.
Let me assume this method is already taught to you in school.

There is also a formula for the y-coordinate of the vertex, but it is not popular, so I assume it was not taught to you in school.
For the y-coordinate of the vertex, we just substitute the value of the x-coordinate in the general form of the equation of the parabola.

The general form of the equation of a vertical parabola is:
y = ax^2 +bx +c ........***

Based on that form, the formula for the x-coordinate of the vertex is -b/(2a).

The axis of symmetry of a vertical parabola is a vertical line that passes through the vertex. Hence, this axis of symmetry is actually,
x = -b/2a. ......***

------------
y=-x^2+2x
y = -x^2 +2x ....(1)

So, the x-coordinate of the vertex is
-b / 2a
= -2 / 2(-1)
= -2 / -2
= 1

Then, substitute that into (1) to get the y-coordinate of the vertex,
y = -x^2 +2x .....(1)
y = -(1)^2 +2(1)
y = -1 +2
y = 1

Therefore, the vertex is (1,1) ...answer.
And the axis of symmetry is the vertical line, x = 1 ...answer.

--------------
y=3x^2+6x-17
y = 3x^2 +6x -17 ....(2)

-b/(2a)
= -6 / 2(3)
= -6 / 6
= -1

y = 3(-1)^2 +6(-1) -17
y = 3(1) -6 -17
y = -20

Therefore, the vertex is (-1,-20). ....answer.
And x = -1 is the axis of symmetry. ....answer.

------------
Let me answer also this third problem, because it is "different" in form.

y=3(x+1)^2 -20
y = 3(x+1)^2 -20

We can rearrange that into
y +20 = 3(x+1)^2 ....(3)
And if you were taught already the standard form of the equation of a vertical parabola:
(y-k) = a(x-h)^2 ....***
Where
(h,k) is the vertex
then, in (3), it is easy to see that
k = -20 and h = -1,
so, the vertex is (-1,-20).
and the axis of symmetry is x = -1.

So let us transform the
y = 3(x+1)^2 -20
into the general form
y = ax^2 +bx +c.

y = 3(x+1)^2 -20
y = 3(x^2 +2x +1) -20
y = 3x^2 +6x +3 -20
y = 3x^2 +6x -17 ....(4)

Ummm, see this Eq.(4).
It is exactly the parabola in the problem just above.
And what are the vertex and axis of symmetry of the same parabola above?