1. **State the problem:** Find the exact location of all relative and absolute extrema of the function $$f(x) = 3x^2 - 12x - 15$$ on the domain $$[0, 5]$$. 2. **Recall the formula and rules:** For a quadratic function $$f(x) = ax^2 + bx + c$$, the vertex occurs at $$x = -\frac{b}{2a}$$. Since $$a = 3 > 0$$, the parabola opens upwards, so the vertex is a minimum point. 3. **Find the vertex:** $$x = -\frac{-12}{2 \times 3} = \frac{12}{6} = 2$$ 4. **Evaluate $$f(x)$$ at the vertex:** $$f(2) = 3(2)^2 - 12(2) - 15 = 3(4) - 24 - 15 = 12 - 24 - 15 = -27$$ 5. **Evaluate $$f(x)$$ at the domain endpoints:** - At $$x=0$$: $$f(0) = 3(0)^2 - 12(0) - 15 = -15$$ - At $$x=5$$: $$f(5) = 3(5)^2 - 12(5) - 15 = 3(25) - 60 - 15 = 75 - 60 - 15 = 0$$ 6. **Determine extrema:** - The vertex at $$x=2$$ is a relative minimum and also the absolute minimum on $$[0,5]$$ because $$f(2) = -27$$ is less than values at endpoints. - The maximum value on the domain is at $$x=5$$ with $$f(5) = 0$$, which is the absolute maximum. 7. **Final answers:** - $$f$$ has a relative minimum and absolute minimum at $$(2, -27)$$. - $$f$$ has an absolute maximum at $$(5, 0)$$.