Question

$\qquad f(x)=x^2-6x+4, f: (-\infty, 3]$

Answer 1

The range of the function $f(x)$ on the domain $(-\infty, 3]$ is $[-5, \infty)$.

Answer 2

The function $f(x)=x^2-6x+4$ is a parabola opening upwards. The vertex is at $x = -(-6)/(2 \times 1) = 3$. The value at the vertex is $f(3) = (3)^2 - 6(3) + 4 = 9 - 18 + 4 = -5$. The domain is $(-\infty, 3]$. On this domain, the function is strictly decreasing as it approaches the vertex from the left. The minimum value is $f(3) = -5$. As $x \to -\infty$, $f(x) \to \infty$. Therefore, the range is $[-5, \infty)$.