Question

worked example 2
the function f is defined by $f(x) = (x - 2)^2 + 3$ for $0 \leqslant x \leqslant 6$.
sketch the graph of the function.
find the range of f.

Explanation:

Step1: Analyze the function type

The function \( f(x)=(x - 2)^2+3 \) is a quadratic function in vertex form \( y = a(x - h)^2 + k \), where \( a = 1 \), \( h = 2 \), \( k = 3 \). Since \( a>0 \), the parabola opens upwards, and the vertex is at \( (2, 3) \).

Step2: Find the minimum value

For a parabola opening upwards, the vertex is the minimum point. So when \( x = 2 \) (which is within the domain \( 0\leq x\leq6 \)), the minimum value of \( f(x) \) is \( f(2)=(2 - 2)^2+3 = 3 \).

Step3: Find the maximum value

We need to check the endpoints of the domain \( x = 0 \) and \( x = 6 \).

• When \( x = 0 \): \( f(0)=(0 - 2)^2+3=4 + 3 = 7 \)
• When \( x = 6 \): \( f(6)=(6 - 2)^2+3=16 + 3 = 19 \)

Since \( 19>7 \), the maximum value of \( f(x) \) on the domain \( 0\leq x\leq6 \) is \( 19 \).

Step4: Determine the range

The range of a function is the set of all possible output values (y - values). Since the minimum value is \( 3 \) and the maximum value is \( 19 \), and the function is continuous on the interval \( [0, 6] \), the range of \( f \) is \( 3\leq f(x)\leq19 \).

Answer:

The range of \( f \) is \( \boldsymbol{3\leq f(x)\leq19} \) (or in interval notation \( [3, 19] \)).