Question 912258
f(x)=10-(x-2)^2


f(x)=-1(x-2)^2 + 10


that equation is in the form y = a(x-h)^2 + k where a = -1, h = 2, k = 10


The value of 'a' is negative, so this is a parabola that opens downward. That means it has a peak point.


The peak is at (2, 10) which is the vertex.


Range: {{{y<=10}}}


Range in interval notation: <img src="http://latex.codecogs.com/gif.latex?(-\infty,&space;10]" title="(-\infty, 10]" />