Answer:
In vertex form the equation is [tex]y=-\frac{1}{2}(x-2)^2+2[/tex]
In standard form the equation is [tex]y=-\frac{1}{2}x^2+2x[/tex]
Step-by-step explanation:
The equation of the directrix tells us that this is an x-squared parabola. Because the directrix is above the vertex, the parabola will open downward. The vertex form of this equation is:
[tex]4p(y-k)=-(x-h)^2[/tex]
where p is the number of units between the directrix and the vertex. The number of units here is .5 or 1/2. Filling in the coordinates of the vertex and the p value of 1/2:
[tex]4(.5)(y-2)=-(x-2)^2[/tex]
Simplifying we have:
[tex]2(y-2)=-(x-2)^2[/tex]
Divide both sides by 2 to get:
[tex]y-2=-\frac{1}{2}(x-2)^2[/tex]
Add 2 to both sides to get the final vertex form:
[tex]y=-\frac{1}{2}(x-2)^2+2[/tex]
If you want that in standard form, you first need to expand the squared term to get:
[tex]y=-\frac{1}{2}(x^2-4x+4)+2[/tex]
Order of operations tells us that we have to distribute in the -1/2 first to get:
[tex]y=-\frac{1}{2}x^2+2x+2-2[/tex]
which simplifies to the standard form:
[tex]y=-\frac{1}{2}x^2+2x[/tex]
