+666 Find the largest real number $c$ such that $1$ is in the range of $f(x)=x^2-5x+c-3x+8$.
+129845 x^2 - 8x + 8 + c
The x coordinate of the vertex is 8/ (2*1) = 4
So....we need y to be 1 when x = 4
1 = (4)^2 - 8(4) + 8 + c
1 = 16 - 32 + 8 + c
1 = -8 + c
c = 9
Here's the graph :
Note that if c is larger than 9, the graph is shited above y =1....so 1 will not be in the range
