1. **Graph the transformed equation**: Given the function $$f(x) = -4 \left( \frac{1}{2} (x + 1) \right)^2 + 5$$. - This is a quadratic function (parabola). - The inner part $$\frac{1}{2}(x+1)$$ compresses the graph horizontally by a factor of 2. - The $$+1$$ inside the parentheses shifts the graph left by 1 unit. - The coefficient $$-4$$ in front reflects the parabola vertically and stretches it vertically by a factor of 4. - The $$+5$$ shifts the graph up by 5 units. The vertex is at $$x = -1$$ and $$y = 5$$. 2. **Solve the radical equation:** $$\sqrt{2m + 10} - \sqrt{m + 6} = 1$$ Step 1: Isolate one radical $$\sqrt{2m + 10} = 1 + \sqrt{m + 6}$$ Step 2: Square both sides $$ (\sqrt{2m + 10})^2 = (1 + \sqrt{m + 6})^2 $$ $$ 2m + 10 = 1 + 2\sqrt{m + 6} + m + 6 $$ Simplify: $$ 2m + 10 = m + 7 + 2\sqrt{m + 6} $$ Rearranged: $$ 2m + 10 - m - 7 = 2 \sqrt{m + 6} $$ $$ m + 3 = 2 \sqrt{m + 6} $$ Step 3: Square both sides again $$ (m + 3)^2 = 4(m + 6) $$ Expand left side: $$ m^2 + 6m + 9 = 4m + 24 $$ Bring all terms to one side: $$ m^2 + 6m + 9 - 4m - 24 = 0 $$ Simplify: $$ m^2 + 2m - 15 = 0 $$ Step 4: Solve quadratic Factor: $$ (m + 5)(m - 3) = 0 $$ Solutions: $$ m = -5 \quad \text{or} \quad m = 3 $$ Step 5: Check for extraneous solutions by plugging back into original equation For $$m = -5$$: $$ \sqrt{2(-5) + 10} - \sqrt{-5 + 6} = \sqrt{0} - \sqrt{1} = 0 - 1 = -1 \neq 1 $$ (Reject) For $$m = 3$$: $$ \sqrt{2(3) + 10} - \sqrt{3 + 6} = \sqrt{16} - \sqrt{9} = 4 - 3 = 1 $$ (Accept) Final solution: $$ m = 3 $$ 3. **Prove the inverse functions:** Given: $$ c(x) = (x + 4)^3 - 9 $$ $$ j(x) = \sqrt[3]{x + 9} - 4 $$ To prove two functions are inverses, show that: $$ c(j(x)) = x $$ and $$ j(c(x)) = x $$ Check $$ c(j(x)) $$: $$ c(j(x)) = \left(j(x) + 4\right)^3 - 9 = \left(\sqrt[3]{x + 9} - 4 + 4\right)^3 - 9 $$ $$ = \left(\sqrt[3]{x + 9}\right)^3 - 9 = x + 9 - 9 = x $$ Check $$ j(c(x)) $$: $$ j(c(x)) = \sqrt[3]{c(x) + 9} - 4 = \sqrt[3]{(x + 4)^3 - 9 + 9} - 4 $$ $$ = \sqrt[3]{(x + 4)^3} - 4 = x + 4 - 4 = x $$ Since both compositions return $$x$$, the functions $$c$$ and $$j$$ are inverses.