Find the largest value of c such that 1 is in the range of f(x)=x^2-5x+c.

xXxTenTacion Aug 24, 2018

#1**+1 **

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

CPhill Aug 24, 2018