1. **Problem statement:** We analyze the extrema of the functions \(f_a(x) = x^2 - ax - 2\), \(g_a(x) = 2ax^3 - 6x\), and \(h_a(x) = -\frac{1}{5}x^3 + \frac{3}{5}ax^2\) depending on parameter \(a\). Then for \(a=1\), find intercepts and extrema. 2. **Formulas and rules:** Extrema occur where the first derivative is zero and the second derivative test determines the type: - If \(f'(x_0) = 0\) and \(f''(x_0) > 0\), minimum. - If \(f'(x_0) = 0\) and \(f''(x_0) < 0\), maximum. 3. **Function \(f_a(x)\):** - Derivative: \(f_a'(x) = 2x - a\) - Set to zero: \(2x - a = 0 \Rightarrow x = \frac{a}{2}\) - Second derivative: \(f_a''(x) = 2 > 0\) always, so minimum at \(x=\frac{a}{2}\) - Minimum value: \(f_a\left(\frac{a}{2}\right) = \left(\frac{a}{2}\right)^2 - a\frac{a}{2} - 2 = \frac{a^2}{4} - \frac{a^2}{2} - 2 = -\frac{a^2}{4} - 2\) 4. **Function \(g_a(x)\):** - Derivative: \(g_a'(x) = 6ax^2 - 6\) - Set to zero: \(6ax^2 - 6 = 0 \Rightarrow x^2 = \frac{1}{a}\) (assuming \(a \neq 0\)) - Critical points: \(x = \pm \frac{1}{\sqrt{a}}\) - Second derivative: \(g_a''(x) = 12ax\) - At \(x = \frac{1}{\sqrt{a}}\), \(g_a'' = 12a \frac{1}{\sqrt{a}} = 12 \sqrt{a} > 0\) minimum - At \(x = -\frac{1}{\sqrt{a}}\), \(g_a'' = -12 \sqrt{a} < 0\) maximum 5. **Function \(h_a(x)\):** - Derivative: \(h_a'(x) = -\frac{3}{5}x^2 + \frac{6}{5}ax\) - Set to zero: \(-\frac{3}{5}x^2 + \frac{6}{5}ax = 0 \Rightarrow x(-3x + 6a) = 0\) - Critical points: \(x=0\) and \(x=2a\) - Second derivative: \(h_a''(x) = -\frac{6}{5}x + \frac{6}{5}a\) - At \(x=0\): \(h_a''(0) = \frac{6}{5}a\) - At \(x=2a\): \(h_a''(2a) = -\frac{6}{5}2a + \frac{6}{5}a = -\frac{12}{5}a + \frac{6}{5}a = -\frac{6}{5}a\) - For \(a>0\): - At \(x=0\), \(h_a''(0) > 0\) minimum - At \(x=2a\), \(h_a''(2a) < 0\) maximum 6. **For \(a=1\):** - \(f_1(x) = x^2 - x - 2\) - Minimum at \(x=\frac{1}{2}\) - Minimum value: \(f_1(\frac{1}{2}) = \frac{1}{4} - \frac{1}{2} - 2 = -\frac{9}{4} = -2.25\) - Intercepts: - \(y\)-intercept: \(f_1(0) = -2\) - \(x\)-intercepts: solve \(x^2 - x - 2 = 0\) \(\Rightarrow (x-2)(x+1)=0\) so \(x=2, -1\) - \(g_1(x) = 2x^3 - 6x\) - Critical points: \(x= \pm 1\) - \(g_1''(x) = 12x\) - At \(x=1\), \(g_1''(1) = 12 > 0\) minimum - At \(x=-1\), \(g_1''(-1) = -12 < 0\) maximum - Intercepts: - \(y\)-intercept: \(g_1(0) = 0\) - \(x\)-intercepts: solve \(2x^3 - 6x = 0 \Rightarrow 2x(x^2 - 3) = 0\) so \(x=0, \pm \sqrt{3}\) - \(h_1(x) = -\frac{1}{5}x^3 + \frac{3}{5}x^2\) - Critical points: \(x=0, 2\) - \(h_1''(x) = -\frac{6}{5}x + \frac{6}{5}\) - At \(x=0\), \(h_1''(0) = \frac{6}{5} > 0\) minimum - At \(x=2\), \(h_1''(2) = -\frac{12}{5} + \frac{6}{5} = -\frac{6}{5} < 0\) maximum - Intercepts: - \(y\)-intercept: \(h_1(0) = 0\) - \(x\)-intercepts: solve \(-\frac{1}{5}x^3 + \frac{3}{5}x^2 = 0 \Rightarrow \frac{1}{5}x^2(-x + 3) = 0\) so \(x=0, 3\) **Final answers:** - \(f_a(x)\) has a minimum at \(x=\frac{a}{2}\) with value \(-\frac{a^2}{4} - 2\). - \(g_a(x)\) has a maximum at \(x=-\frac{1}{\sqrt{a}}\) and minimum at \(x=\frac{1}{\sqrt{a}}\). - \(h_a(x)\) has a minimum at \(x=0\) and maximum at \(x=2a\). - For \(a=1\), intercepts and extrema as detailed above.