- #1

stephensnoah1

- 1

- 0

y^2+12y+16x+68=0

The form we have been using is (y-k)^2=4p(x-h)

Any explanation would help too, I'm really stuck on this one.

Thank you!

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In summary, the equation of the given parabola can be rewritten in the form (y-k)^2=4p(x-h), with k=-6, h=-32, and p=-4. This means the vertex is located at (-32, -6), the focus is at (-33, -6), and the directrix is the vertical line x=-30. To find these values, we completed the square by adding and subtracting the appropriate constant and rearranging the terms. The leading coefficient of -1/16 indicates that the parabola opens to the left and never crosses the y-axis.

- #1

stephensnoah1

- 1

- 0

y^2+12y+16x+68=0

The form we have been using is (y-k)^2=4p(x-h)

Any explanation would help too, I'm really stuck on this one.

Thank you!

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- #2

Olinguito

- 239

- 0

$$16x\ =\ 68-12y-y^2\ =\ 104-(y+6)^2$$

whence

$$(y+6)^2\ =\ 104-16x\ =\ 4(-4)\left(x-\frac{13}2\right)$$

in the form you have been using.

- #3

HOI

- 921

- 2

In other words "complete the square".

- #4

jonah1

- 107

- 0

When I saw this a month ago, I plotted it with my graphing app along with the OP's y^2+12y+16x+68=0. They didn't match. I figured somebody will come along with a correction but nobody did. Forgot about it till I decided to remove a few open tabs and saw it again. I guess nobody here is no longer double checking.Olinguito said:$$(y+6)^2\ =\ 104-16x\ =\ 4(-4)\left(x-\frac{13}2\right)$$

That should have been

$$(y+6)^2=4(-4)(x+2)$$

Edit: Just realized something else. This thread is a year old already. Country Boy's post made me think it was just new a month ago.

Last edited:

- #5

HOI

- 921

- 2

A "perfect square" is of the form $(y+ a)^2= y^2+ 2ay+ a^2$. Comparing $y^2+ 2ay$ with $y^2+ 12y$ we must have $2ay= 12y$ so a= 6 and then $a^2= 36$. We need to add 36 to make this a "perfect square".

We can't just add a number to an expression and have the same numerical value but we can add and subtract the same number: $y^2+ 12y= y^2+ 12y+ 36-36= (y+ 6)^2- 36$.

So $y^2+12y+16x+68= (y+ 6)^2- 36+ 16x+ 68= (y+ 6)^2+ 16x+ 32= 0$.

$16x= -(y+6)^2- 32$

$x= -\frac{1}{16}(y+ 6)^2- 32$.

Now, that is a parabola with horizontal axis (parallel to the x-axis), opening to the left. Since a square in never negative, that is $-32$ minus something. x will be largest when $(y+ 6)^2= 0$, y= 6, where x= -32. The vertex is at (-32, -6).

When y= 0, $x= -\frac{1}{16}(0- 6)^2- 32= -\frac{9}{4}- 32= -34.25$. The x- intercept is at (-34.25, 0).

Since the leading coefficient, $-\frac{1}{16}$, is negative, the parabola opens to the left and never crosses the y- axis. There is no y- intercept.

- #6

jonah1

- 107

- 0

That should have beenCountry Boy said:$x= -\frac{1}{16}(y+ 6)^2- 32$.

$x= -\frac{1}{16}(y+ 6)^2- \frac{32}{16}$

- #7

HOI

- 921

- 2

And $\frac{32}{16}= 2$! Yes, thank you.

- #8

jonah1

- 107

- 0

It need not be said but this of course invalidates some the rest afterCountry Boy said:And $\frac{32}{16}= 2$! Yes, thank you.

$x= -\frac{1}{16}(y+ 6)^2- 32$

Cheers.

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The vertex of a parabola is the point where the parabola changes direction and is located at the minimum or maximum point of the parabola. To find the vertex, we can use the formula `x = -b/2a`

where `a`

and `b`

are the coefficients of the quadratic terms in the equation. In this case, the vertex is located at `(-2, -6)`

.

The focus of a parabola is a point inside the parabola that is equidistant from the directrix and the vertex. To find the focus, we can use the formula `F = (-b/2a, c - (b^2 - 1)/4a)`

where `F`

is the focus, `a`

and `b`

are the coefficients of the quadratic terms, and `c`

is the constant term in the equation. In this case, the focus is located at `(-2, -7.75)`

.

The directrix of a parabola is a line that is perpendicular to the axis of symmetry and is located on the opposite side of the vertex from the focus. To determine the directrix, we can use the formula `x = -b/2a + p`

where `p`

is the distance between the vertex and the directrix. In this case, the directrix is the line `x = 2.25`

.

Yes, we can graph the parabola to visualize the position of the vertex, focus, and directrix. By plotting the points we calculated for the vertex and focus, and drawing the directrix, we can see that the parabola opens downwards and the vertex is located at the lowest point on the parabola.

The vertex, focus, and directrix of a parabola can be used to solve various real-world problems, such as finding the minimum or maximum value of a function, determining the optimal point in a business or engineering problem, or predicting the trajectory of a projectile. By understanding the properties of a parabola and how to find its key points, we can apply this knowledge to solve practical problems in different fields.

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