1. **Problem 2a:** Express $2x^2 - 8x + 5$ in the form $a(x + b)^2 + c$, where $a$, $b$, and $c$ are integers. 2. The formula to complete the square for a quadratic $ax^2 + bx + c$ is: $$a\left(x + \frac{b}{2a}\right)^2 + \left(c - a\left(\frac{b}{2a}\right)^2\right)$$ 3. For $2x^2 - 8x + 5$, identify $a=2$, $b=-8$, and $c=5$. 4. Calculate $\frac{b}{2a} = \frac{-8}{2 \times 2} = \frac{-8}{4} = -2$. 5. Write the expression as: $$2\left(x - 2\right)^2 + \left(5 - 2 \times (-2)^2\right)$$ 6. Calculate the constant term inside the parentheses: $$5 - 2 \times 4 = 5 - 8 = -3$$ 7. So the expression in completed square form is: $$2(x - 2)^2 - 3$$ --- 1. **Problem 2b:** Write down the equation of the line of symmetry for the graph of $y = 2x^2 - 8x + 1$. 2. The line of symmetry for a parabola $y = ax^2 + bx + c$ is given by: $$x = -\frac{b}{2a}$$ 3. Here, $a=2$, $b=-8$, so: $$x = -\frac{-8}{2 \times 2} = \frac{8}{4} = 2$$ 4. Therefore, the line of symmetry is: $$x = 2$$ --- 1. **Problem 3a:** Express $7 + 5x - x^2$ in the form $a - (x + b)^2$, where $a$ and $b$ are constants. 2. Rewrite the quadratic in standard form: $$-x^2 + 5x + 7$$ 3. Factor out $-1$ from the $x^2$ and $x$ terms: $$-(x^2 - 5x) + 7$$ 4. Complete the square inside the parentheses: $$x^2 - 5x = x^2 - 5x + \cancel{\left(\frac{5}{2}\right)^2} - \cancel{\left(\frac{5}{2}\right)^2} = (x - \frac{5}{2})^2 - \left(\frac{5}{2}\right)^2$$ 5. Substitute back: $$-\left[(x - \frac{5}{2})^2 - \left(\frac{5}{2}\right)^2\right] + 7 = - (x - \frac{5}{2})^2 + \left(\frac{5}{2}\right)^2 + 7$$ 6. Calculate constants: $$\left(\frac{5}{2}\right)^2 = \frac{25}{4}$$ 7. So: $$- (x - \frac{5}{2})^2 + \frac{25}{4} + 7 = - (x - \frac{5}{2})^2 + \frac{25}{4} + \frac{28}{4} = - (x - \frac{5}{2})^2 + \frac{53}{4}$$ 8. Express in the form $a - (x + b)^2$: $$a = \frac{53}{4}, \quad b = -\frac{5}{2}$$ --- 1. **Problem 3b:** Find the coordinates of the turning point of $y = 7 + 5x - x^2$ and state whether it is a maximum or minimum. 2. From the completed square form: $$y = - (x - \frac{5}{2})^2 + \frac{53}{4}$$ 3. The turning point is at $x = \frac{5}{2}$. 4. Substitute $x = \frac{5}{2}$ into $y$: $$y = - (\frac{5}{2} - \frac{5}{2})^2 + \frac{53}{4} = \frac{53}{4}$$ 5. Since the coefficient of the squared term is negative ($-1$), the parabola opens downward, so the turning point is a maximum. 6. Therefore, the turning point is: $$\left(\frac{5}{2}, \frac{53}{4}\right)$$ which is a maximum point.