Question 975422
y = -(x-2)²+7
<pre>
You memorize the general vertex form

y = a(x-h)²+k

You compare your equation to that:

a=-1, h=2, k=7

The vertex is the point (h,k) = (2,7)

You plot that point:

{{{drawing(400,400,-3,7,-2,8,

graph(400,400,-3,7,-2,8),

circle(2,7,0.11),circle(2,7,0.09),circle(2,7,0.07),circle(2,7,0.05),circle(2,7,0.03),circle(2,7,0.01) )}}}

Now from that vertex, go right and left 1 unit and then "a" units.
When a is positive you go up, and when it's negative you go down.
So we go right and left 1 unit from the vertex and down 1 unit from
each because "a" is -1.

{{{drawing(400,400,-3,7,-2,8,

graph(400,400,-3,7,-2,8),


circle(3,6,0.09),circle(3,6,0.07),circle(3,6,0.05),circle(3,6,0.03),circle(3,6,0.01),

circle(1,6,0.09),circle(1,6,0.07),circle(1,6,0.05),circle(1,6,0.03),circle(1,6,0.01),
circle(2,7,0.11),circle(2,7,0.09),circle(2,7,0.07),circle(2,7,0.05),circle(2,7,0.03),circle(2,7,0.01) )}}}


Then we sketch in the graph:

{{{drawing(400,400,-3,7,-2,8,

graph(400,400,-3,7,-2,8,-(x-2)^2+7),


circle(3,6,0.09),circle(3,6,0.07),circle(3,6,0.05),circle(3,6,0.03),circle(3,6,0.01),

circle(1,6,0.09),circle(1,6,0.07),circle(1,6,0.05),circle(1,6,0.03),circle(1,6,0.01),
circle(2,7,0.11),circle(2,7,0.09),circle(2,7,0.07),circle(2,7,0.05),circle(2,7,0.03),circle(2,7,0.01) )}}}

Edwin</pre>