Question

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

Answer 1

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 2

Answer:

$y=a(x-h)^{2} +k$

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.