Question

Find the focus of the parabola y = x^2 + 5x

Answer 1

\[y = x ^{2}+5x\]

Answer 2

Somehow, you have to express it in the form
\[\Large (x-\color{red}a)^2 =4p(y-\color{blue}b)\]

Answer 3

We need to complete the square.... do you know how to do that? :)

Answer 4

yes

Answer 5

Well then, complete the square for the x-part

Answer 6

\[\left( x ^{2} + 2.25\right)^{2} = y+6.25\]

Answer 7

well, not x^2 in the parentheses

Answer 8

What exactly does the p stand for?

Answer 9

hang on...

Answer 10

Where did 2.25 come from

Answer 11

2.5 is what I meant

Answer 12

Oh, okay, great :)
So...
\[\Large \left(x +\frac52\right)^2 = y+\frac{25}{4}\]
(I used fractions, but it's essentially the same)

Answer 13

Okay, so about that p value... that's actually the key-value here :)
You need to find the value of p, so... use this equation...
\[\Large \left(y+\frac{25}4\right)=4\cdot \frac14 \left(y+\frac{25}4\right)\]

Answer 14

that part clear? ^

Answer 15

Yes, that makes sense

Answer 16

So, rewrite...
\[\Large \left(x +\frac52\right)^2 =4\cdot \frac14\left( y+\frac{25}{4}\right)\]

Answer 17

What do we do with the 1/4 now?

Answer 18

That is your p-value.

So, you now want to find the vertex of your parabola.
Anyway, now that you have completed the square, it shouldn't be hard....
\[\Large y = \left(x+\frac52\right)^2-\frac{25}4\]
What's the vertex?

Answer 19

-5/2?

Answer 20

Vertex. The whole vertex, and not just the x-value of the vertex.
The vertex is the point
\[\Large \left(-\frac52 \ , \ -\frac{25}4\right)\]

Answer 21

After we find the vertex, how do we find the focus?

Answer 22

Sorry, I think I was very vague when guiding you through this...
Three things we need to find the focus...
Vertex (check)
p-value (check)
concavity

Which way does this parabola open?

Answer 23

Upwards

Answer 24

Right. So we now have all we need to find the focus.

the FOCUS is the point that is exactly p-units away from the vertex in the UPWARD direction :)

Answer 25

So , -24/4 = -6. Thank you very much! Just one more question. How do you determine if it is concave up or concave down? (I had an image of the graph that I looked at for this one).

Answer 26

Wait, -6 is NOT the focus, that's just the y-value of the focus... just sayin' the focus is
\[\Large \left(-\frac52 \ , \ -6\right)\]
but other than that, good job :)

Answer 27

Now, the concavity is determined by the coefficient of \(\large x^2\).
If it's negative, then the parabola opens down. If it's positive, then the parabola opens up

Answer 28

haha, thanks for keeping me straight

Answer 29

Okay, great. Thanks so much!