I have opened the image. Let's go through the questions one by one and solve them:

Question 1.1

Find the slope-intercept form of the equation of the line that passes through the point $(5, -3)$ and is perpendicular to the line $y = 2x + 1$.

Solution: The slope of the line $y = 2x + 1$ is $m = 2$. For two lines to be perpendicular, the product of their slopes must be $-1$. Hence, the slope of the required line is: $m_{\text{perpendicular}} = -\frac{1}{2}$ Now, use the point-slope form of the equation $y - y_1 = m(x - x_1)$ with point $(5, -3)$: $y + 3 = -\frac{1}{2}(x - 5)$ Simplifying to slope-intercept form: $y = -\frac{1}{2}x + \frac{5}{2} - 3$ $y = -\frac{1}{2}x - \frac{1}{2}$ So, the equation is: $y = -\frac{1}{2}x - \frac{1}{2}$

Question 1.2

Find the solution of the equation $x^2 - 2x - 3 = 0$ by graphing.

Solution: The equation can be factored as: $x^2 - 2x - 3 = (x - 3)(x + 1) = 0$ Thus, the solutions are $x = 3$ and $x = -1$. These are the points where the parabola intersects the x-axis.

Question 1.3

Compute $\lim_{x \to 6} \frac{\sqrt{x + 3} - 3}{x - 6}$.

Solution: We use the technique of multiplying by the conjugate: $\lim_{x \to 6} \frac{\sqrt{x+3} - 3}{x - 6} \times \frac{\sqrt{x+3} + 3}{\sqrt{x+3} + 3}$ $= \lim_{x \to 6} \frac{(x+3) - 9}{(x-6)(\sqrt{x+3} + 3)} = \lim_{x \to 6} \frac{x - 6}{(x - 6)(\sqrt{x+3} + 3)}$ $= \lim_{x \to 6} \frac{1}{\sqrt{x+3} + 3} = \frac{1}{\sqrt{6+3} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

Question 1.4

Find the instantaneous rate of change of $f(x) = x^3 + 1$ at $x = 2$ using first principles.

Solution: The instantaneous rate of change is the derivative of $f(x)$, given by: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ Substitute $f(x) = x^3 + 1$ and compute: $f'(x) = \lim_{h \to 0} \frac{(x+h)^3 + 1 - (x^3 + 1)}{h} = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}$ $= \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2) = 3x^2$ At $x = 2$: $f'(2) = 3(2^2) = 12$

Question 1.5

Solve $x^2 = 5x - 7$ using the quadratic formula.

Solution: Rearrange the equation: $x^2 - 5x + 7 = 0$ Using the quadratic formula: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(7)}}{2(1)} = \frac{5 \pm \sqrt{25 - 28}}{2} = \frac{5 \pm \sqrt{-3}}{2}$ The discriminant is negative, so the solutions are complex: $x = \frac{5 \pm i\sqrt{3}}{2}$ Thus, the solutions are $x = \frac{5}{2} \pm \frac{i\sqrt{3}}{2}$.

If you'd like me to proceed with the remaining questions, or if you need any clarifications, feel free to ask! Here are five follow-up questions:

1. How do you interpret the slope of a line that is perpendicular to another?
2. What do the solutions of the quadratic equation $x^2 - 2x - 3 = 0$ represent geometrically?
3. Why is multiplying by the conjugate useful in limit problems?
4. How is the derivative linked to the instantaneous rate of change?
5. What does a negative discriminant in a quadratic equation tell you about the nature of the solutions?

Tip: When solving quadratic equations, always check the discriminant to quickly assess the nature of the roots.