Aparabola with a vertex at (0,0) has a directrix that crosses the negative part of the y-axis. which could be the equation of the parabola? x2 = –4y x2 = 4y y2 = 4x y2 = –4x

In the attached files you can see graphs of all parabolas and their directrices.

As you can see the graph of parabola with a vertex at (0,0) and with a directrix that crosses the negative part of the y-axis is represented at the second diagram.

correct choice is B,

Step-by-step explanation:

Since, if the directrix of a parabola crosses the negative part of the y-axis.

Then it is obvious that the parabola must be upward,

Since, the equation of upward parabola is,

Where, a > 1,

Where (h, k) is the vertex of the parabola,

h = 0, k = 0,

Then the equation of the upward parabola,

∵ a > 0 ⇒ 1/a > 0,

Since, -4 < 0,

Hence, the required equation of the parabola would be,

Note : and have the directrix on the negative x-axis and positive x-axis respectively.

answer

see attachment

explanation

the given function is

this function has y-intercept at

-4 and a slope of 2.

note that, the graph has a positive slope.

the graph that represents this function is the straight line that moves upward from left to right.

the graph touches the y-axis at -4.

the correct choice is the graph shown in the attachment.

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.