Derivative as Slope

The derivative is the slope!

Finding the Tangent Line Equation

Once you know the slope $m = f'(a)$ and the point $(a, f(a))$, use point-slope form:

$y - f(a) = f'(a)(x - a)$

Or rearrange to slope-intercept form:

$y = f'(a)(x - a) + f(a)$

Example 1: Find the Tangent Line

Find the equation of the tangent line to $f(x) = x^2$ at $x = 2$.

Step 1: Find the point $f(2) = 2^2 = 4$ Point: $(2, 4)$

Step 2: Find the slope (derivative) Using the power rule (we'll learn this soon): $f'(x) = 2x$ $f'(2) = 2(2) = 4$ Slope: $m = 4$

Step 3: Use point-slope form $y - 4 = 4(x - 2)$ $y - 4 = 4x - 8$ $y = 4x - 4$

Answer: The tangent line is $y = 4x - 4$

Positive, Negative, and Zero Slopes

The derivative tells you the curve's behavior:

Critical Points

When $f'(a) = 0$, we have a critical point:

• Could be a maximum (top of a hill)
• Could be a minimum (bottom of a valley)
• Could be a saddle point (neither)

The horizontal tangent line is a clue that something interesting is happening!

Example 2: Horizontal Tangent

Find where $f(x) = x^2 - 4x + 3$ has a horizontal tangent line.

Step 1: Find the derivative $f'(x) = 2x - 4$

Step 2: Set equal to zero $2x - 4 = 0$ $2x = 4$ $x = 2$

Step 3: Find the point $f(2) = 4 - 8 + 3 = -1$

Answer: Horizontal tangent at $(2, -1)$

This is actually the vertex (minimum) of the parabola!

Secant Lines vs. Tangent Lines

Secant line: Connects two points on the curve

• Slope: $\frac{f(b) - f(a)}{b - a}$ (average rate of change)

Tangent line: Touches at one point

• Slope: $f'(a)$ (instantaneous rate of change)

As the two points of a secant line get closer, the secant line approaches the tangent line!

Normal Lines

The normal line is perpendicular to the tangent line.

If the tangent line has slope $m = f'(a)$, the normal line has slope:

$m_{\text{normal}} = -\frac{1}{f'(a)}$

(Negative reciprocal)

Graphical Analysis

Looking at a graph:

• Steep tangent → Large $|f'(a)|$ → Fast change
• Gentle tangent → Small $|f'(a)|$ → Slow change
• Horizontal tangent → $f'(a) = 0$ → No change (momentarily)

Practice Visualization

For any curve:

1. Pick a point
2. Imagine a line that "just kisses" the curve there
3. That line's slope is the derivative
4. If you can't draw it without lifting your pencil, it's not differentiable there!

Step 2: Set equal to zero $3x^2 - 6x = 0$

Step 3: Factor $3x(x - 2) = 0$

Step 4: Solve $x = 0 \text{ or } x = 2$

Step 5: Find the y-coordinates

• At $x = 0$: $g(0) = 0 - 0 = 0$
• At $x = 2$: $g(2) = 8 - 12 = -4$

Answer: Horizontal tangent lines at $(0, 0)$ and $(2, -4)$

These are the critical points where the function changes from increasing to decreasing or vice versa!