```Question 425905
How to find the vertx, value of p, axis of symmetry, focus, and directrix of each parabola, and then graph.
y= (1/32)x^2
hOW TO write the equation in standard for for each porabola
a. vertex (0,0), focus(0,1)
b. vertex(0,0), focus (-8,0)

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y=1/32(x^2
This of the form, (x-h)^2=4p(y-k), (h,k) being the (x,y) coordinates of the vertex. Since you don't see h or k in the equation, the vertex must be at the origin, (0,0).
rewriting given equation,
x^2=32y
This is a parabola that opens upwards with its vertex at the (0,0), and axis of symmetry,x=0 or the y-axis.
4p=32
p=8
The focus on the axis of symmetry is 8 units above the center at (0,8).
The directrix is a line, y=-8
See the graph of this parabola below:
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{{{ graph( 200, 200, -10, 10, -10, 10, (1/32)x^2,-8) }}}

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a. vertex (0,0), focus(0,1)
From the focal point (0,1), it can be seen that the axis of symmetry is the y-axis like the previous parabola and it opens upwards with p=1 and its directrix at y=-1
Equation: x^2=4y
See the graph below:(Note that the higher the coefficient of x^2, the steeper the curve, that is, a higher slope)
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{{{ graph( 200, 200, -5, 5, -5, 5, x^2/4,-1) }}}
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b. vertex(0,0), focus (-8,0)
Again, gleaning information from the focal point, it can be seen that the axis of symmetry is x=0 or the x-axis and it opens sideways to the left with p=8 and its directrix at x=8.
Equation:
y^2=-32x
See the graph below:
{{{ graph( 200, 200, -10, 10, -10, 10,(-32x)^.5,-(-32x)^.5) }}}
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