Question:

Step 1: For the range of $f(x)$ to be $[0, \infty)$, $f(x)$ must represent an upward-opening parabola with a minimum value of $0$. This requires the leading coefficient to be positive: $$k - 3 > 0 \Rightarrow k > 3$$ Step 2: The minimum value of a quadratic function $Ax^2+Bx+C$ occurs at $x = -\frac{B}{2A}$. For $f(x) = (k-3)x^2 - 2kx + 3k - 6$, the x-coordinate of the vertex is: $$x = \frac{-(-2k)}{2(k-3)} = \frac{2k}{2(k-3)} = \frac{k}{k-3}$$ Step 3: The minimum value of $f(x)$ is obtained by substituting this x-coordinate into the function: $$f\left(\frac{k}{k-3}\right) = (k-3)\left(\frac{k}{k-3}\right)^2 - 2k\left(\frac{k}{k-3}\right) + 3k - 6$$ Step 4: Simplify the expression for the minimum value: $$f\left(\frac{k}{k-3}\right) = \frac{k^2}{k-3} - \frac{2k^2}{k-3} + 3k - 6$$ $$f\left(\frac{k}{k-3}\right) = -\frac{k^2}{k-3} + 3k - 6$$ $$f\left(\frac{k}{k-3}\right) = -\frac{k^2}{k-3} + 3(k-2)$$ Step 5: For the range to be $[0, \infty)$, the minimum value must be $0$. $$-\frac{k^2}{k-3} + 3(k-2) = 0$$ Multiply by $(k-3)$ (which is non-zero since $k>3$): $$-k^2 + 3(k-2)(k-3) = 0$$ $$3(k-2)(k-3) = k^2$$ Step 6: Expand and rearrange the equation: $$3(k^2 - 5k + 6) = k^2$$ $$3k^2 - 15k + 18 = k^2$$ $$2k^2 - 15k + 18 = 0$$ Step 7: Solve the quadratic equation for $k$ using the quadratic formula $k = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $$k = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(2)(18)}}{2(2)}$$ $$k = \frac{15 \pm \sqrt{225 - 144}}{4}$$ $$k = \frac{15 \pm \sqrt{81}}{4}$$ $$k = \frac{15 \pm 9}{4}$$ This yields two possible values for $k$: $$k_1 = \frac{15 + 9}{4} = \frac{24}{4} = 6$$ $$k_2 = \frac{15 - 9}{4} = \frac{6}{4} = \frac{3}{2}$$ Step 8: Apply the condition from Step 1, $k > 3$. The value $k = \frac{3}{2}$ does not satisfy $k > 3$. The value $k = 6$ satisfies $k > 3$. Therefore, the only valid value for $k$ is $6$.