how do you find the vertex of these parabolas?

Question

To find the vertex please use the completing the square formula and show how you did it.


x^2 - 2x + 44y + 353 = 0

x^2 + 6x + 8y + 25 = 0


and how do you write these parabolas in the form of y = ax+ b

-48x = y^2

4y^2 = -48x

PLEASE show how you did it.

Answer 1

1. a.) The vertex form for a parabola is:

y = a (x - h)² + k, where (h, k) is the vertex of the parabola and c is a constant. There is also another form used, y - k = a (x - h)², but it is essentially the same as the first, with k transposed to the other side of the equation. I will use the first form here, since it is more direct.

Let's get the first equation into the form y = a (x - h)² + k.

44y = - x² + 2x - 353
44y = - (x² - 2x) - 353

We need to complete the trinomial square on the right side. We do that by taking half the coefficient of the linear x term, which is -1, then we square it and add that result to both sides of the equation, since we have to keep the equation balanced.

44y - (-2/1)² = - [x² - 2x + (-2/1)²] - 353
44y - 1 = - x² + 2x - 1 - 353
44y - 1 = - (x - 1)² - 353
44y = - (x - 1)² - 353 + 1
44y = - (x - 1)² - 352

Notice this. When I transposed the terms containing the x variables to the right side of the equation, their signs changed to the opposite of what they were before. I factored -1 out of both of them to make it easier to complete the trinomial square, because to do that, the coefficient of the leading term needs to be 1 only, not -1 or any other number. Now, when I add the complementary term, 1, to the right side, I need to compensate for the fact that I factored out the -1 before. To do that, I redistribute the -1 and the +1 which I added becomes -1. That means I actually subtracted 1 from the right side, so that means I must also subtract 1 from the left hand side of the equation to keep the equation balanced. That's why you see 1 being subtracted from the left side of the equation. Then we proceed to get the equation back into the proper form. But we still need to do more to actually find k. To do that, we need the coefficient of the y containing variable to be 1. To do that, we simply divide both sides of the equation by 44.


44y/44 = [-(x - 1)² - 352] /44
y = -1/44(x - 1)² - 8

Now we have our equation in the final form. Remember, our vertex is at (h, k). Since x - h = x - 1, then h = +1. Of course, we know k = -8 because it appears as such in the equation, and a = -1/44.

So this parabola has vertex at (1, -8) and opens downward because c < 0.


1. b.) We do something similar with this equation. First get it in the right form. But we still can't tell what h and k are. So we will have to work with it:

x² + 6x + 8y + 25 = 0
8y = -x² - 6x - 25

In the next step we complete the trinomial square, again factoring out -1 from the x variable terms before doing so for the same reason as the first problem. We see again that we are subtracting a number from the right side, so we must also subtract the same number from the left side. Then we proceed to get the equation into the final form:

8y - (6/2)² = - [x² + 6x + (6/2)²] - 25
8y - 9 = - (x² + 6x + 9) - 25
8y - 9 = - (x + 3)² - 25
8y = - (x + 3)² - 25 + 9
8y = - (x + 3)² - 16
8y/8 = -(x + 3)²/8 - 16/8
y = -1/8(x + 3)² - 2

x - h = x + 3, so h = -3 and k = -2

This one has vertex at (-3,-2) with a = -1/8. So this parabola also opens downward.

2.) Are you sure you have this written correctly? I have never heard of a parabola being written in the form y = ax + b. That is a form for a linear equation, and a parabola is a quadratic. A parabola can be written in the form y = ax² + b, where b is both the y-intercept and y coordinate of the vertex of the parabola.

y = ax² + b is actually a morph of the vertex equation given above, i.e. y = ax² + b is equivalent to y = a(x + 0)² + b, where the vertex is located at (0, b). Then you could write these equations like this:

a.) -48x = y² ----> x = - (1/48)y² + 0 = - (1/48)(y - 0)² + 0, with vertex at (0, 0) and represents a parabola which opens to the left, since a > 0, and has axis of symmetry which is the x-axis.

b.) 4y² = - 48x.
Divide through by - 48 first. Then this is the result:

- (1/12)y² = x ----> x = - (1/12)y², which we can also write:

x = - (1/12)(y - 0)² + 0. This parabola also has vertex at (0, 0), with axis of symmetry equal to the x-axis. But it opens to the left, instead of the right because a < 0.

Answer 2

x^2-2x+44y+353
=(x^2-2x+1)+44y -1+353(for completing square)
=(x-1)^2+44y+352=0
The equation for a parabola is:
(x-h)^2=4p(y-k) (vertex at h,k)
Rewriting the above equation,:
(x-1)^2=-44y-352=-44(y+8)
=-44{(y-(-8)};h=1,k=-8(to get standard form)
Vertex is at(1,-8)


x^2+6x+8y+25=X^2+6x+9+8y-9+25
=(x+3)^2+8y+16=0
So (x+3)^2=-8y-16
{x-(-3)]^2=-8(y+2)
=-8{y-(-2)}
h=-3,k=-2
Vertex is (-3,-2)

Answer 3

i can tell you what I remember

first set up the equation to be easily completed

x^2 - 2x _ + 44y = -353
Take the -2 / by 2 and square it
x^2 - 2x +1 + 44y = -353
Add 1 to both sides of the equation to balance it and factor
(x-1) +44y = -352

Answer 4

ok. y = x^2-4x+3 and you desire to locate the vertex. it is a ordinary variety for a parabola the place y = ax^2 +bx + c. If a > 0 then the parabola has a minimum and is concave upwards. vertex is placed at -b/2a and y(-b/2a) so on your case -b/2a = 4/2 = 2 and y(2) = -a million So, your vertex is at (2,-a million)