1. parabolas

x = -1/2y squared - y - 4

x = -1/2y squared - y - 4

i need the vertex, x intercept, y intercepts and directrix...that would be marvelous.

3. Originally Posted by brittneyv
oops i forgot about that part.
sorry, i need the vertex, directrix, x-axis and y-axis.
Re-write the equation into the vertex-directrix-form:

$\displaystyle x = -\frac12 y^2 - y - 4 ~\iff~-2x = y^2 + 2y +1-1+8 ~\iff~-2\left(x+\frac72 \right) = (y+1)^2~\iff~$ $\displaystyle 2 \cdot (-1)\cdot \left(x+\frac72 \right) = (y+1)^2$

The vertex has the coordinates $\displaystyle V\left(-\frac72 , -1 \right)$

Therefore the directrix has the equation d: $\displaystyle x = -\frac72 -\left(-\frac12 \right) = -3$

With your need for x-axis and y-axis I assume that you mean the x and y intercepts of the parabola:

x-intercept: y = 0:

$\displaystyle 2 \cdot (-1)\cdot \left(x+\frac72 \right) = (0+1)^2~\iff~ x = -4$

y-intercept: x = 0:

$\displaystyle 2 \cdot (-1)\cdot \left(0+\frac72 \right) = (y+1)^2$ The LHS is negative, the RHS is a square - but a square can't be negative and therefore there doesn't exist any y-intercept.

Another attempt: The vertex is left of the y-axis and the parabola opens to left. Therefore the parabola and the y-axis can't have a common point. Thus: No y-intercept.