Question

Help?

Answer 1

Do you know the forumla to find the vertex and focus?

Answer 2

@johnweldon1993 ???

Answer 3

So there are quite a few shortcuts with parabolas in standard form....if we convert this to standard form we have

\[\large y = \frac{1}{8}x^2 - \frac{3}{4}x + \frac{49}{8}\]

The vertex (h,k) can be found using the equations \(\large h= -\frac{b}{2a}\) \(\large k = \frac{4ac - b^2}{4a}\)
The focus, you first find 'p' using \(\large p = \frac{1}{4a}\) then use that in the formula \[\large F(h, k + p)\]
And the directrix is just found by using \(\large y = k - p\)

So lets see Vertex, you should have gotten \(\large h = -\frac{b}{2a} = -\frac{-\frac{3}{4}}{2 \times \frac{1}{8}} = -\frac{-\frac{3}{4}}{\frac{1}{4}} = 3\)
And \(\large k = \frac{4\times \frac{1}{8} \times \frac{49}{8} - (-\frac{3}{4})^2}{4\times \frac{1}{8}} = 5\) Making the Vertex (3,5) right?

Answer 4

Yeah I got the vertex thanks! its (3,5)

Answer 5

I didnt know how to find the directix and the focus... mind explaining???

Answer 6

Now on to the Focus...I have the equation of \(\large F(h,k + p)\)

We have 'h' and 'k' we just need p

\[\large p = \frac{1}{4a} = \frac{1}{4\times \frac{1}{8}} = \frac{1}{\frac{1}{2}} = 2\]

So we would have the focus as \(\large (h,k + p) = (3, 5 + 2 = (3,7)\)

Answer 7

I still dont get it.. hang on.

Answer 8

Of course :)

But to sum it up, like I said there are a bunch of shortcuts for finding these things.
The Focus coordinates would be at the point (h,k + p)

where \(\large h = -\frac{b}{2a}\) \(\large k = \frac{4ac - b^2}{4a}\) and \(\large p = \frac{1}{4a}\)

We just went through the formulas :)

Answer 9

I got it... lol so we basically gave ourselves the answer earlier.. so the directix is 3?

Answer 10

Correct! :)

Answer 11

y=3*

Answer 12

Lol I knew what you meant :P