# What is the image of the function f(x)=x^2-4x+3 for x>=3?What is the image of the function f(x)=x^2-4x+3 for x>=3?

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### 2 Answers

You need to evaluate the interval that contains the images of the given function, for `x >= 3` , such that:

`x^2 - 4x + 3 = y => x^2 - 4x = y - 3`

You need to complete the square to the left side, using the following formula, such that:

`(a - b)^2 = a^2 - 2ab + b^2`

Reasoning by analogy yields:

`x^2 - 4x + 4 = y - 3 + 4 => (x - 2)^2 = y + 1`

`x - 2 = +-sqrt(y + 1) => x = 2 +- sqrt(y + 1)`

The problem provides the information that `x>=3` , such that:

`{(x>=3),(x = 2 +- sqrt(y + 1)):} => 2 +- sqrt(y + 1) >= 3`

`+- sqrt(y + 1) >= 3 - 2 => +- sqrt(y + 1) >= 1`

`sqrt(y + 1) >= 1 => y + 1 >= 1 => y >= 0`

`- sqrt(y + 1) >= 1 => sqrt(y + 1) <= -1` invalid

**Hence, evaluating the constraint for values of y to be the images of the given function, under the given conditions, yields **`y >= 0.`

We'll put y = x^2-4x+3

We'll subtract y both sides:

x^2 - 4x + 3 - y = 0

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

x1 = [4 + sqrt(16 - 4(3-y))]/2

x2 = [4 - sqrt(16 - 4(3-y))]/2

But, from enunciation, we know that x>=3

[4 + sqrt(16 - 4(3-y))]/2 >=3

We'll multiply by 2:

[4 + sqrt(16 - 4(3-y))] >= 6

We'll subtract 4 both sides:

sqrt(16 - 4(3-y)) >= 2

We'll raise to square both sides:

16 - 4(3-y) >= 4

We'll subtract 16 both sides:

- 4(3-y) >= 4 - 16

- 4(3-y) >= -12

We'll divide by -4:

3-y =< 3

y >= 0

The image of the function is represented by the range [0 , +infinite), for x >=3.