Question

Quadratics: Writing Vertex Form, Algebra 1- Mrs. Daniel, Kuta Software

Answer 1

`y=a(x-h)^(2) +k`

Where
`(h,k)` 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
`(-4,1)`, and a points that is on the parabola could be
`(-3,2)`. So, with this information, we can find the
`a` parameter and form the parabola equation. Replacing all these, we have

`y=a(x-h)^(2) +k`

`2=a(-3-(-4))^(2) +1`

Then, we solve for
`a`

`2=a(-3+4)^(2)+1\\ 2=a(1)^(2)+1\\ 2=a+1\\2-1=a\\a=1`

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

`y=1(x-(-4))^(2) +1\\ \therefore y=(x+4)^(2)+1`

2.

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

`y=a(x-h)^(2) +k`

Replacing all values, that is,
`x=3`,
`y=-3`,
`h=2` and
`k=-4`.

`-3=a(3-2)^(2) +(-4)\\-3=a(1)^(2) -4\\-3+4=a\\a=1`

So, the equation of this parabola is

`y=(x-2)^(2) -4`

3.

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

`y=a(x-h)^(2) +k\\2=a(4-3)^(2)+3\\2-3=a\\a=-1`

Observe that the parameter
`a` is negative, that indicates the parabola is downside.

So, the equation is

`y=a(x-h)^(2) +k\\y=-1(x-3)^(2) +3\\y=-(x-3)^(2) +3`

4.

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

`y=a(x-h)^(2) +k\\-2=a(-2-(-1))^(2) +(-1)\\-2=a(-2+1)^(2) -1\\-2+1=a\\a=-1`

So, the equation is
`y=-(x+1)^(2) -1`

5.

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

`y=a(x-h)^(2) +k\\4=a(2-1)^(2) +2\\4-2=a\\a=2`

The equation would be

`y=a(x-h)^(2) +k\\y=2(x-1)^(2) +2`

6.

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

`y=a(x-h)^(2) +k\\0=a(2-3)^(2)+(-2)\\ 2=a(-1)^(2) \\a=2`

So, the equation is

`y=a(x-h)^(2) +k\\y=2(x-3)^(2) -2`

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

Answer 2

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