Functions: Question 1

Worked solution

Step 1: Identify the shape of the graph

$h(x) = (x-4)^2 + 1$ is an upward-opening parabola. Ignoring the domain restriction for a moment, its vertex, the minimum point of the full parabola, is at $(4, 1)$.

Step 2: Consider the effect of the domain restriction

The domain is restricted to $x \le 4$, so we only ever see the left-hand branch of the parabola, the branch that runs into the vertex from the left.

For $x \le 4$:

• As $x$ increases towards $4$, the term $(x-4)^2$ decreases towards $0$, so $h(x)$ decreases towards its smallest value $h(4) = (4-4)^2+1 = 1$.
• As $x$ decreases further (moves further left, $x \to -\infty$), $(x-4)^2 \to \infty$, so $h(x) \to \infty$.

So on this restricted domain, $h$ is a strictly decreasing function of $x$ (equivalently, strictly increasing as $x$ moves away from $4$), and it takes every value from $1$ upward, with no upper bound.

Step 3: State the range

The smallest value of $h$ occurs at the domain boundary $x=4$, where $h(4)=1$, and $h$ grows without bound as $x$ decreases. So the range is: $h(x) \ge 1$

Why the other options are wrong

• B ($h(x)\le 1$): reverses the inequality. $h$ takes values from $1$ upward, not downward.
• C ($h(x)\ge 4$): mistakes the domain bound $x\le4$ for a bound on the output values.
• D ($h(x)\le 4$): combines both errors above.

Final answer

$\boxed{h(x) \ge 1} \quad \text{(Option A)}$