+586 Find the largest value of c such that 1 is in the range of f(x)=x^2-5x+c.
+129899 Since this parabola turns upward....1 will still be in the range when, at a minimum, (m , 1) is the
vertex
We have the form ax^2 + bx + c
To find "m" we can use -b/ [2a] = 5 / (2 * 1) = 5/2 = m
So (5/2, 1) is the vertex
So....subbing this into the function for x and f(x) to find " c," we have
1 = (5/2)^2 - 5(5/2) + c
1 = 25/4 - 25/2 + c
1 = 25/4 - 50/4 + c
1 = -25/4 + c
1 + 25/4 = c
29/4 = c
Here's a graph to show this : https://www.desmos.com/calculator/gbhwg6nerl
Check for yourself that if c > 29/4, then 1 will be out of range
