1. **Expand and simplify:** Given $f(x) = -3(x - 2)(x + 4)$ Use the distributive property (FOIL) to expand: $$(x - 2)(x + 4) = x^2 + 4x - 2x - 8 = x^2 + 2x - 8$$ Multiply by $-3$: $$f(x) = -3(x^2 + 2x - 8) = -3x^2 - 6x + 24$$ 2. **Write the function in vertex form:** Given $f(x) = x^2 - 8x + 5$ Use completing the square: $$f(x) = x^2 - 8x + 5 = (x^2 - 8x + 16) - 16 + 5 = (x - 4)^2 - 11$$ So vertex form is: $$f(x) = (x - 4)^2 - 11$$ 3. **Identify vertex, axis of symmetry, direction of opening:** For $f(x) = -2x^2 + 4x + 6$ Vertex formula: $$h = -\frac{b}{2a} = -\frac{4}{2(-2)} = 1$$ Find $k$: $$k = f(1) = -2(1)^2 + 4(1) + 6 = -2 + 4 + 6 = 8$$ Vertex: $(1, 8)$ Axis of symmetry: $x = 1$ Direction of opening: Since $a = -2 < 0$, parabola opens downward. 4. **Find maximum or minimum value:** For $f(x) = x^2 - 10x + 21$ Vertex $h = -\frac{b}{2a} = -\frac{-10}{2(1)} = 5$ Calculate $k$: $$k = f(5) = 5^2 - 10(5) + 21 = 25 - 50 + 21 = -4$$ Since $a=1 > 0$, parabola opens upward, so minimum value is $-4$ at $x=5$. 5. **Discriminant when graph touches x-axis at exactly one point:** Discriminant formula: $$\Delta = b^2 - 4ac$$ If the graph touches x-axis once, the quadratic has one real root, so: $$\Delta = 0$$ 6. **Find equation of parabola opening downward with vertex $(1,8)$ and y-intercept 6:** Vertex form: $$f(x) = a(x - h)^2 + k = a(x - 1)^2 + 8$$ Use y-intercept $(0,6)$: $$6 = a(0 - 1)^2 + 8 = a(1) + 8$$ Solve for $a$: $$a + 8 = 6 \Rightarrow a = -2$$ Equation: $$f(x) = -2(x - 1)^2 + 8$$