Question

how would you limit the domain to make this function one to one and still have the same range?
$f(x) = (x - 2)^2 + 7$
$x \geq ?$

Explanation:

Step1: Analyze the function type

The function \( f(x)=(x - 2)^2+7 \) is a quadratic function in vertex form \( f(x)=a(x - h)^2 + k \), where \( a = 1 \), \( h = 2 \), and \( k = 7 \). The graph of a quadratic function \( y=a(x - h)^2 + k \) is a parabola. Since \( a=1>0 \), the parabola opens upwards.

Step2: Find the vertex and axis of symmetry

The vertex of the parabola \( y=a(x - h)^2 + k \) is \( (h,k) \), so the vertex of \( f(x)=(x - 2)^2+7 \) is \( (2,7) \). The axis of symmetry of the parabola \( y=a(x - h)^2 + k \) is the vertical line \( x = h \), so the axis of symmetry for this function is \( x = 2 \).

Step3: Determine the domain for one - to - one

For a parabola that opens upwards, the function is decreasing to the left of the axis of symmetry and increasing to the right of the axis of symmetry. A function is one - to - one if it passes the horizontal line test. To make the function \( f(x)=(x - 2)^2+7 \) one - to - one while keeping the same range, we can consider the part of the function where it is either increasing or decreasing. Since the range of \( f(x)=(x - 2)^2+7 \) is \( y\geq7 \) (because the square term \( (x - 2)^2\geq0 \), so \( (x - 2)^2+7\geq7 \)), and we want to keep the same range, we can take the domain as \( x\geq2 \) (the part of the function to the right of the axis of symmetry, where the function is increasing). If we take \( x\leq2 \), the range will still be \( y\geq7 \), but the question asks for \( x\geq[?] \), so we choose the increasing part.

Answer:

\( 2 \)