1. (a) Express $16x^2 - 24x + 10$ in the form $(4x + a)^2 + b$. Start with the expression: $$16x^2 - 24x + 10$$ Factor out 16 from the first two terms: $$16\left(x^2 - \frac{24}{16}x\right) + 10 = 16\left(x^2 - \frac{3}{2}x\right) + 10$$ Complete the square inside the parentheses. Take half of $-\frac{3}{2}$, which is $-\frac{3}{4}$, and square it: $$\left(-\frac{3}{4}\right)^2 = \frac{9}{16}$$ Add and subtract $\frac{9}{16}$ inside the parentheses: $$16\left(x^2 - \frac{3}{2}x + \frac{9}{16} - \frac{9}{16}\right) + 10 = 16\left(\left(x - \frac{3}{4}\right)^2 - \frac{9}{16}\right) + 10$$ Distribute 16: $$16\left(x - \frac{3}{4}\right)^2 - 16 \times \frac{9}{16} + 10 = 16\left(x - \frac{3}{4}\right)^2 - 9 + 10$$ Simplify constants: $$16\left(x - \frac{3}{4}\right)^2 + 1$$ So, $$(4x - 3)^2 + 1$$ Therefore, $a = -3$ and $b = 1$. 2. (b) Given $16x^2 - 24x + 10 = k$ has exactly one root, find the root. For a quadratic equation to have exactly one root, the discriminant must be zero: $$\Delta = b^2 - 4ac = 0$$ Here, $a=16$, $b=-24$, and $c=10 - k$ (since $16x^2 - 24x + 10 = k$ rearranges to $16x^2 - 24x + (10 - k) = 0$). Calculate discriminant: $$(-24)^2 - 4 \times 16 \times (10 - k) = 0$$ Simplify: $$576 - 64(10 - k) = 0$$ Distribute: $$576 - 640 + 64k = 0$$ Simplify: $$-64 + 64k = 0$$ Divide both sides by 64: $$\cancel{64}k - \cancel{64} = 0 \Rightarrow k - 1 = 0$$ So, $$k = 1$$ Substitute $k=1$ back into the equation: $$16x^2 - 24x + 10 = 1 \Rightarrow 16x^2 - 24x + 9 = 0$$ Use the form from part (a): $$(4x - 3)^2 + 1 = 1 \Rightarrow (4x - 3)^2 = 0$$ Solve for $x$: $$4x - 3 = 0 \Rightarrow x = \frac{3}{4}$$ 3. (b) Describe the transformation from $y = x^2$ to $y = x^2 + 6x + 5$. Rewrite $y = x^2 + 6x + 5$ by completing the square: Take half of 6, which is 3, square it to get 9: $$y = x^2 + 6x + 9 - 9 + 5 = (x + 3)^2 - 4$$ This shows the graph of $y = x^2$ is shifted left by 3 units and down by 4 units. 4. Find $k$ and the point of tangency for the line $y = kx - k$ tangent to $y = -\frac{1}{2x}$. Set the line equal to the curve at the tangent point $x = t$: $$k t - k = -\frac{1}{2t}$$ Rewrite: $$k t - k + \frac{1}{2t} = 0$$ The slope of the curve at $x = t$ is the derivative: $$y = -\frac{1}{2x} = -\frac{1}{2} x^{-1}$$ Derivative: $$y' = -\frac{1}{2} \times (-1) x^{-2} = \frac{1}{2x^2}$$ The slope of the tangent line is $k$, so: $$k = \frac{1}{2 t^2}$$ From the line equation at $x = t$: $$k t - k = -\frac{1}{2 t}$$ Substitute $k = \frac{1}{2 t^2}$: $$\frac{1}{2 t^2} t - \frac{1}{2 t^2} = -\frac{1}{2 t}$$ Simplify left side: $$\frac{1}{2 t} - \frac{1}{2 t^2} = -\frac{1}{2 t}$$ Multiply both sides by $2 t^2$ to clear denominators: $$2 t^2 \times \left(\frac{1}{2 t} - \frac{1}{2 t^2}\right) = 2 t^2 \times \left(-\frac{1}{2 t}\right)$$ Calculate each term: $$t - 1 = -t$$ Add $t$ to both sides: $$t - 1 + t = 0 \Rightarrow 2t - 1 = 0$$ Solve for $t$: $$2t = 1 \Rightarrow t = \frac{1}{2}$$ Find $k$: $$k = \frac{1}{2 t^2} = \frac{1}{2 \times (\frac{1}{2})^2} = \frac{1}{2 \times \frac{1}{4}} = \frac{1}{\frac{1}{2}} = 2$$ Find $y$ coordinate: $$y = k t - k = 2 \times \frac{1}{2} - 2 = 1 - 2 = -1$$ So the tangent point is at $\left(\frac{1}{2}, -1\right)$ and $k = 2$.