1. **Determine the roots and x-value of the vertex for each function:** **a.** $f(x) = (x-2)$ - This is a linear function, not quadratic, so it has one root at $x=2$. - No vertex for linear functions. **b.** $g(x) = 2x^2 + 2x - 12$ - Roots found by solving $2x^2 + 2x - 12 = 0$. - Use quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4(2)(-12)}}{2 \times 2} = \frac{-2 \pm \sqrt{4 + 96}}{4} = \frac{-2 \pm \sqrt{100}}{4} = \frac{-2 \pm 10}{4}$$ - Roots: $$x_1 = \frac{-2 - 10}{4} = \frac{-12}{4} = -3$$ $$x_2 = \frac{-2 + 10}{4} = \frac{8}{4} = 2$$ - Vertex x-value: $$x = -\frac{b}{2a} = -\frac{2}{2 \times 2} = -\frac{2}{4} = -0.5$$ **c.** $h(x) = 7x^2$ - Roots: Set $7x^2 = 0 \Rightarrow x=0$ - Vertex x-value: $$x = -\frac{b}{2a} = -\frac{0}{2 \times 7} = 0$$ **d.** About the roots and vertex x-value: - For a quadratic function $ax^2 + bx + c$, the vertex x-value is always halfway between the roots. - This is because vertex x-value is $-\frac{b}{2a}$ and roots sum to $-\frac{b}{a}$, so vertex x-value is half the sum of roots. - Therefore, the correct statement is: **The x-value of the vertex is halfway between the roots.** 2. Given $f(x) = (x-7)(x+6)$: **a.** Solve for $x$ when $f(x) = 0$: - Set $(x-7)(x+6) = 0$. - Roots: $x=7$ or $x=-6$. **b.** Roots are $7$ and $-6$. **c.** x-value producing max or min is halfway between roots: $$x = \frac{7 + (-6)}{2} = \frac{1}{2} = 0.5$$ **d.** Find max or min value by evaluating $f(0.5)$: $$f(0.5) = (0.5 - 7)(0.5 + 6) = (-6.5)(6.5) = -42.25$$ - Since coefficient of $x^2$ is positive (1), this is a minimum. 3. Determine roots: **a.** $f(x) = (3x + 11)(x - 4)$ - Roots: $3x + 11 = 0 \Rightarrow x = -\frac{11}{3}$ - $x - 4 = 0 \Rightarrow x = 4$ **b.** $g(x) = x^2 - 6^2 = x^2 - 36$ - Roots: $x^2 = 36 \Rightarrow x = \pm 6$ **c.** $h(x) = 5x^2 - 42x + 12$ - Use quadratic formula: $$x = \frac{42 \pm \sqrt{(-42)^2 - 4(5)(12)}}{2 \times 5} = \frac{42 \pm \sqrt{1764 - 240}}{10} = \frac{42 \pm \sqrt{1524}}{10}$$ - Simplify $\sqrt{1524} = \sqrt{4 \times 381} = 2\sqrt{381}$ - Roots: $$x = \frac{42 \pm 2\sqrt{381}}{10} = \frac{42}{10} \pm \frac{2\sqrt{381}}{10} = 4.2 \pm 0.2\sqrt{381}$$ **d.** $j(x) = 6x^2 + 14x - 4$ - Quadratic formula: $$x = \frac{-14 \pm \sqrt{14^2 - 4(6)(-4)}}{2 \times 6} = \frac{-14 \pm \sqrt{196 + 96}}{12} = \frac{-14 \pm \sqrt{292}}{12}$$ - Simplify $\sqrt{292} = \sqrt{4 \times 73} = 2\sqrt{73}$ - Roots: $$x = \frac{-14 \pm 2\sqrt{73}}{12} = -\frac{14}{12} \pm \frac{2\sqrt{73}}{12} = -\frac{7}{6} \pm \frac{\sqrt{73}}{6}$$ 4. Determine vertex: **a.** $f(x) = (4x - 11)(10x - 6)$ - Expand: $$f(x) = 40x^2 - 44x - 60x + 66 = 40x^2 - 104x + 66$$ - Vertex x-value: $$x = -\frac{b}{2a} = -\frac{-104}{2 \times 40} = \frac{104}{80} = 1.3$$ - Vertex y-value: $$f(1.3) = 40(1.3)^2 - 104(1.3) + 66 = 40(1.69) - 135.2 + 66 = 67.6 - 135.2 + 66 = -1.6$$ **b.** $g(x) = x^2 - 8^2 = x^2 - 64$ - Vertex at $x=0$ (since no $x$ term) - Vertex y-value: $$g(0) = 0 - 64 = -64$$ **c.** $h(x) = 3x^2 + 12x + 4$ - Vertex x-value: $$x = -\frac{12}{2 \times 3} = -\frac{12}{6} = -2$$ - Vertex y-value: $$h(-2) = 3(-2)^2 + 12(-2) + 4 = 3(4) - 24 + 4 = 12 - 24 + 4 = -8$$ 5. Rock thrown upward: $f(t) = -16t^2 + 32t + 138$ **a.** Bridge height is initial height at $t=0$: $$f(0) = 138$$ **b.** Rock hits water when height is 0: $$-16t^2 + 32t + 138 = 0$$ - Divide by -2 for simpler numbers: $$8t^2 - 16t - 69 = 0$$ - Use quadratic formula: $$t = \frac{16 \pm \sqrt{(-16)^2 - 4(8)(-69)}}{2 \times 8} = \frac{16 \pm \sqrt{256 + 2208}}{16} = \frac{16 \pm \sqrt{2464}}{16}$$ - Simplify $\sqrt{2464} = \sqrt{16 \times 154} = 4\sqrt{154}$ - Roots: $$t = \frac{16 \pm 4\sqrt{154}}{16} = 1 \pm \frac{\sqrt{154}}{4}$$ - Positive root (time after throw): $$t = 1 + \frac{\sqrt{154}}{4} \approx 1 + \frac{12.41}{4} = 1 + 3.10 = 4.10$$ **c.** Time to max height is vertex x-value: $$t = -\frac{b}{2a} = -\frac{32}{2 \times (-16)} = -\frac{32}{-32} = 1$$ **d.** Max height: $$f(1) = -16(1)^2 + 32(1) + 138 = -16 + 32 + 138 = 154$$ 6. Quadratic with roots $x=-3$ and $x=9$, and $f(0)=13$: **a.** $x_1 = -3$, $x_2 = 9$ **b.** Write $f(x) = c(x + 3)(x - 9)$ - Use $f(0) = 13$: $$13 = c(0 + 3)(0 - 9) = c(3)(-9) = -27c$$ - Solve for $c$: $$c = -\frac{13}{27}$$ **c.** Function formula: $$f(x) = -\frac{13}{27}(x + 3)(x - 9)$$ - Expand: $$f(x) = -\frac{13}{27}(x^2 - 6x - 27) = -\frac{13}{27}x^2 + \frac{78}{27}x + 13$$ 7. Rock thrown from 58 feet above road, max height at $t=0.79$, hits road at $t=3.06$. - General form: $$f(t) = at^2 + bt + c$$ - Known: $c = 58$ (height at $t=0$) - Vertex at $t=0.79$ means: $$t_{vertex} = -\frac{b}{2a} = 0.79$$ - Hits road at $t=3.06$ means: $$f(3.06) = 0 = a(3.06)^2 + b(3.06) + 58$$ From vertex formula: $$b = -2a(0.79) = -1.58a$$ Substitute into $f(3.06) = 0$: $$a(3.06)^2 + (-1.58a)(3.06) + 58 = 0$$ $$a(9.3636) - 4.8348a + 58 = 0$$ $$a(9.3636 - 4.8348) = -58$$ $$a(4.5288) = -58$$ $$a = -\frac{58}{4.5288} \approx -12.81$$ Find $b$: $$b = -1.58a = -1.58(-12.81) = 20.23$$ Function: $$f(t) = -12.81t^2 + 20.23t + 58$$