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Mathematics 14 Online
OpenStudy (anonymous):

Medal and fan given for answer! Determine the vertex, focus, and directrix of the parabola without graphing. See the equation below:

OpenStudy (anonymous):

\[y+(x+2)^2=3\]

OpenStudy (anonymous):

Vertex is easy, how much is y when x=-2

OpenStudy (anonymous):

3

OpenStudy (anonymous):

So what is the vertex?

OpenStudy (anonymous):

(-2,3)

OpenStudy (anonymous):

Excellent

OpenStudy (anonymous):

How do you find the focus now?

OpenStudy (anonymous):

Is the parabola concave up or concave down?

OpenStudy (anonymous):

Focus is (0,p) or (p,0) so I need to find p. It is concave down.

OpenStudy (anonymous):

No not (0,p), it is (-2,p)

OpenStudy (anonymous):

Oops, that makes sense since it's translated.

OpenStudy (anonymous):

\[ 4 \frac 1 4(y-3) =- (x+2)^2 \] How much is p?

OpenStudy (anonymous):

How did you get 4 (1/4) ?

OpenStudy (anonymous):

To find p the equation must be \[ 4 p(y-y_0)=(x-x_0)^2 \] we have 1 in front of (y-3) so \[ 1 = 4 \times \frac 14 \]

OpenStudy (anonymous):

How much is p?

OpenStudy (anonymous):

1/4

OpenStudy (anonymous):

ok, since the parabola is concave down, where is the focus?

OpenStudy (anonymous):

(-2,1/4)

OpenStudy (anonymous):

No,\[ (-2,3-\frac 14)\]

OpenStudy (anonymous):

(-2, 11/4)

OpenStudy (anonymous):

The directrix is y= 3+1/4=13/4

OpenStudy (anonymous):

The focus must be on the axis of the parabola

OpenStudy (anonymous):

Did you understand it?

OpenStudy (anonymous):

Yes! Thanks so much.

OpenStudy (anonymous):

YW

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