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do the following expressions and calculations

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### **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.

A set of math problems related to slope-intercept form, limits, derivatives, integrals, and quadratic equations.

This set of questions covers advanced topics in algebra and calculus, including finding slope-intercept forms, solving quadratic equations, computing limits, and using differentiation rules. Perfect for Grades 11-12 students preparing for exams.

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