# Determine the image of the function f(x)=x^2-6x+5 for x>=4

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f(x) = x^2-6x+5.

To determine the image of the function for x> 4..

f(x) = x^2-6x+5.

f(x) = x^2-5x -x+5.

f() = x(x-5)-1(x-5).

Threfore f(x) = (x-1)(x-5).

f(x) is a continous funtion.

Therefore the image f(x) < 0 for x belonging to the interval (1,5) and f(x) > 0, when x < 1, Or when x>5.

Therefore the image f(x) < 0 for 4 =< x < 5 , f(x) < 0.

The image f(x) = 0 when x =5.

And f(x) > 0 , for x > 5.

x f(x) = (x-1)(x-5)= x^2-6x+5.

4 -3

4.5 -1.75,

5 0

5.5 2.25

6 5.

We'll put y = x^2-6x+5

We'll subtract y both sides:

x^2 - 6x + 5 - y = 0

We'll determine the roots of the equtaion with quadratic formula:

x1 = [6 + sqrt(36 - 4(5-y))]/2

x2 = [6 - sqrt(36 - 4(5-y))]/2

But, from enunciation, we know that x>=4

[6 + sqrt(36 - 4(5-y))]/2 >=4

We'll multiply by 2:

6 + sqrt(36 - 4(5-y) >= 8

We'll subtract 6 both sides:

sqrt(36 - 4(5-y)) >= 2

We'll raise to square both sides:

36 - 4(5-y) >= 4

We'll subtract 36 both sides:

- 4(5-y) >= 4 - 36

- 4(5-y) >= -32

We'll divide by -4:

5-y =< 8

y >= -3

**The image of the function belongs to the range [-3 , +infinite), for x >=4.**