11. Solve the following. In finding derivative, use the defintion. 1. Find the slope of the tangent line to the graph of the furction $$ y=x^{2}-3 x+4 $$ at $$ (1,-2) $$. Wrile the equation of the tangent line. Moreover, Find and solue the slope and equation of the Normal line 2. Find the slope of the tangent line to the graph of the function $$ y=x^{2}-5 $$ at $$ (2,3) $$. Wrile the equation of the tangent line. Moreover, Find and solve the slope and eavation of. the normal line.

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Sure, let's solve both problems step-by-step using the definition of the derivative.

### Problem 1

**Function:** $$ y = x^2 - 3x + 4 $$  
**Point of Tangency:** $$ (1, 2) $$ *(Note: At $$ x = 1 $$, $$ y = 1^2 - 3(1) + 4 = 2 $$, so the correct point is $$ (1, 2) $$ instead of $$ (1, -2) $$.)*

#### a) Finding the Slope of the Tangent Line

The derivative of $$ y $$ with respect to $$ x $$ gives the slope of the tangent line.

Using the definition of the derivative:
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
$$

For $$ f(x) = x^2 - 3x + 4 $$:
$$
f(x+h) = (x+h)^2 - 3(x+h) + 4 = x^2 + 2xh + h^2 - 3x - 3h + 4
$$
$$
f(x+h) - f(x) = (x^2 + 2xh + h^2 - 3x - 3h + 4) - (x^2 - 3x + 4) = 2xh + h^2 - 3h
$$
$$
f'(x) = \lim_{h \to 0} \frac{2xh + h^2 - 3h}{h} = \lim_{h \to 0} (2x + h - 3) = 2x - 3
$$

At $$ x = 1 $$:
$$
f'(1) = 2(1) - 3 = -1
$$
So, the **slope of the tangent line** at $$ (1, 2) $$ is **-1**.

#### b) Equation of the Tangent Line

Using the point-slope form:
$$
y - y_1 = m(x - x_1)
$$
Where $$ m = -1 $$ and $$ (x_1, y_1) = (1, 2) $$:
$$
y - 2 = -1(x - 1) \implies y = -x + 1 + 2 \implies y = -x + 3
$$
So, the **tangent line equation** is:
$$
y = -x + 3
$$

#### c) Slope and Equation of the Normal Line

The **normal line** is perpendicular to the tangent line. Therefore, its slope is the **negative reciprocal** of the tangent slope.

$$
m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{-1} = 1
$$

Using the point-slope form with $$ m = 1 $$ and $$ (1, 2) $$:
$$
y - 2 = 1(x - 1) \implies y = x - 1 + 2 \implies y = x + 1
$$
So, the **normal line equation** is:
$$
y = x + 1
$$

---

### Problem 2

**Function:** $$ y = x^2 - 5 $$  
**Point of Tangency:** $$ (2, -1) $$ *(Note: At $$ x = 2 $$, $$ y = 2^2 - 5 = -1 $$, so the correct point is $$ (2, -1) $$ instead of $$ (2, 3) $$.)*

#### a) Finding the Slope of the Tangent Line

Using the definition of the derivative:
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
$$

For $$ f(x) = x^2 - 5 $$:
$$
f(x+h) = (x+h)^2 - 5 = x^2 + 2xh + h^2 - 5
$$
$$
f(x+h) - f(x) = (x^2 + 2xh + h^2 - 5) - (x^2 - 5) = 2xh + h^2
$$
$$
f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x
$$

At $$ x = 2 $$:
$$
f'(2) = 2(2) = 4
$$
So, the **slope of the tangent line** at $$ (2, -1) $$ is **4**.

#### b) Equation of the Tangent Line

Using the point-slope form:
$$
y - y_1 = m(x - x_1)
$$
Where $$ m = 4 $$ and $$ (x_1, y_1) = (2, -1) $$:
$$
y - (-1) = 4(x - 2) \implies y + 1 = 4x - 8 \implies y = 4x - 8 - 1 \implies y = 4x - 9
$$
So, the **tangent line equation** is:
$$
y = 4x - 9
$$

#### c) Slope and Equation of the Normal Line

The **normal line** is perpendicular to the tangent line. Therefore, its slope is the **negative reciprocal** of the tangent slope.

$$
m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{4}
$$

Using the point-slope form with $$ m = -\frac{1}{4} $$ and $$ (2, -1) $$:
$$
y - (-1) = -\frac{1}{4}(x - 2) \implies y + 1 = -\frac{1}{4}x + \frac{1}{2} \implies y = -\frac{1}{4}x + \frac{1}{2} - 1 \implies y = -\frac{1}{4}x - \frac{1}{2}
$$
So, the **normal line equation** is:
$$
y = -\frac{1}{4}x - \frac{1}{2}
$$

---

### Summary

**Problem 1:**
- **Slope of Tangent Line at (1, 2):** $$ -1 $$
- **Tangent Line Equation:** $$ y = -x + 3 $$
- **Slope of Normal Line:** $$ 1 $$
- **Normal Line Equation:** $$ y = x + 1 $$

**Problem 2:**
- **Slope of Tangent Line at (2, -1):** $$ 4 $$
- **Tangent Line Equation:** $$ y = 4x - 9 $$
- **Slope of Normal Line:** $$ -\frac{1}{4} $$
- **Normal Line Equation:** $$ y = -\frac{1}{4}x - \frac{1}{2} $$
**Problem 1:** - **Slope of Tangent Line at (1, 2):** -1 - **Tangent Line Equation:** $$ y = -x + 3 $$ - **Slope of Normal Line:** 1 - **Normal Line Equation:** $$ y = x + 1 $$ **Problem 2:** - **Slope of Tangent Line at (2, -1):** 4 - **Tangent Line Equation:** $$ y = 4x - 9 $$ - **Slope of Normal Line:** -1/4 - **Normal Line Equation:** $$ y = -\frac{1}{4}x - \frac{1}{2} $$

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