Respuesta :

CastleRook
vertex form is given by:
y=a(x-h)^2+k
where the vertex is (h,k)
7. (h,k)=(-4,1)
plugging in the equation we get:
y=a(x+4)^2+1
but substituting (0,2) in the equation and solving for a we get:
2=a(0+4)^2+1
a=1/16
hence:
Answer: y=1/16(x+4)^2+1

8] 
(h,k)=(2,-4)
thus
y=a(x-2)^2-4
plugging point (3,0) in the eqn and solving for a we get
0=a(3-2)^2-4
0=a-4
a=4
hence;
Answer: y=a(x-2)^2-4

9] (h,k)=(3,3)
thus;
y=a(x-3)^2+3
plugging (2,2) in the equation we get:
2=a(-1)^2+3
a=-1
thus;
Answer: y=-1(x-3)^2+3

10] (h,k)=(-1,-1)
y=a(x+1)^2-1
plugging (0,-3) in the equation and solving for a we get:
-3=a(1)^2-1
a=-2
thus
Answer: y=-2(x+1)^2-1

11] (h,k)=(1,2)
y=a(x-1)^2+2
plugging (0,4) in the equation and solving for a we get:
4=a(-1)^2+2
a=2
thus
y=2(x-1)^2+2

12] (h,k)=(3,-2)
y=a(x-3)^2-2
plugging (2,0) and solving for a we get:
0=a(2-3)^2-2
a=2
thus
t=2(x-3)^2-2

Answer:

[tex]y=a(x-h)^{2} +k[/tex]

Where [tex](h,k)[/tex] represents the coordinates of the vertex, the points where the parabola returns.

1.

In the first exercise (number 7 in the image), we observe that the vertex is at [tex](-4,1)[/tex], and a points that is on the parabola could be [tex](-3,2)[/tex]. So, with this information, we can find the [tex]a[/tex] parameter and form the parabola equation. Replacing all these, we have

[tex]y=a(x-h)^{2} +k[/tex]

[tex]2=a(-3-(-4))^{2} +1[/tex]

Then, we solve for [tex]a[/tex]

[tex]2=a(-3+4)^{2}+1\\ 2=a(1)^{2}+1\\ 2=a+1\\2-1=a\\a=1[/tex]

Now we have all the parameters, the equation of the first parabola is

[tex]y=1(x-(-4))^{2} +1\\ \therefore y=(x+4)^{2}+1[/tex]

2.

In the second exercies (number 8 in the image), the vertex or returning point is at [tex](2,-4)[/tex], and one point on the parabola is (3,-3). Doing the same process as we did in the first exercise, we have

[tex]y=a(x-h)^{2} +k[/tex]

Replacing all values, that is, [tex]x=3[/tex], [tex]y=-3[/tex], [tex]h=2[/tex] and [tex]k=-4[/tex].

[tex]-3=a(3-2)^{2} +(-4)\\-3=a(1)^{2} -4\\-3+4=a\\a=1[/tex]

So, the equation of this parabola is

[tex]y=(x-2)^{2} -4[/tex]

3.

(Number 9 in the image). We apply the same process here. The vertex is at [tex](3,3)[/tex], one point on the parabola is [tex](4,2)[/tex]. Then, we replace in the explicit form to find [tex]a[/tex]

[tex]y=a(x-h)^{2} +k\\2=a(4-3)^{2}+3\\2-3=a\\a=-1[/tex]

Observe that the parameter [tex]a[/tex] is negative, that indicates the parabola is downside.

So, the equation is

[tex]y=a(x-h)^{2} +k\\y=-1(x-3)^{2} +3\\y=-(x-3)^{2} +3[/tex]

4.

(Number 10 in the image). Vertex at (-1,-1), one point on the parabola is (-2,-2). Replacing

[tex]y=a(x-h)^{2} +k\\-2=a(-2-(-1))^{2} +(-1)\\-2=a(-2+1)^{2} -1\\-2+1=a\\a=-1[/tex]

So, the equation is [tex]y=-(x+1)^{2} -1[/tex]

5.

(Number 11 in the image). Vertex at (1,2) and point at (2,4). Replacing

[tex]y=a(x-h)^{2} +k\\4=a(2-1)^{2} +2\\4-2=a\\a=2[/tex]

The equation would be

[tex]y=a(x-h)^{2} +k\\y=2(x-1)^{2} +2[/tex]

6.

(Number 12 in the image). Vertex at (3,-2), point at (2.0). Replacing in the explicit form, we have

[tex]y=a(x-h)^{2} +k\\0=a(2-3)^{2}+(-2)\\ 2=a(-1)^{2} \\a=2[/tex]

So, the equation is

[tex]y=a(x-h)^{2} +k\\y=2(x-3)^{2} -2[/tex]

So, there you have all equations of each parabola. The process in the same for all of them.