Question 1201701
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NOTE for future reference: the symbol "^" (shift-6) is commonly used to represent exponents.  So you can write the equation for this problem as<br>
12y=(x-1)^2-48
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The x term is squared, so the graph opens up or down.  The general vertex form of the equation of a parabola I prefer to use is this:<br>
{{{y-k=(1/(4p))(x-h)^2}}}<br>
Note many references will show this equation in different equivalent forms; and different students have different preferences on which form to use.  Some common equivalent forms are<br>
{{{y=(1/(4p))(x-h)^2+k}}}  [puts only "y" on the left]<br>
{{{4p(y-k)=(x-h)^2}}}  [keeps the "4p" on the left so it is not a fraction]<br>
In any of those forms, the vertex is (h,k); p is the directed distance (i.e., can be negative) from the directrix to the vertex and from the vertex to the focus.<br>
Put the equation in your example in this form:<br>
{{{12y=(x-1)^2-48}}}<br>
{{{12y+48=(x-1)^2}}}<br>
{{{12(y+4)=(x-1)^2}}}<br>
{{{y+4=(1/12)(x-1)^2}}}<br>
{{{y-(-4)=(1/(4*3))(x-1)^2}}}<br>
The vertex is (1,-4) and p is 3.<br>
The directrix is p = 3 units below the vertex, at y = -7.<br>
The focus is p = 3 units above the vertex, at (1,-1).<br>
ANSWERS:
vertex (1,-4)
focus (1,-1)
directrix y = -7<br>