The tangent line

Three quick questions from earlier lessons.

The blue line is the tangent to $y = x^2$ at $x = a$ . Move $a$ .

The parabola y = x² and its tangent line at x = a, which touches the curve at (a, a²) with slope 2a.

At $a = 1$ , the tangent has slope 2 and touches the curve at $(1, 1)$ . Which equation is it?

The tangent line touches the graph at $(a, f(a))$ with slope $f'(a)$ .

Tangent line to

y = f(x)

at

x = a

:

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

.

$f(2) = 5$ and $f'(2) = -3$ . Write the tangent line at $x = 2$ .

y = 5 + (-3)(x - 2)

The point (2, 5) and the slope −3.

y = -3x + 11

Expand if you like.

Your turn: $f(4) = 1$ and $f'(4) = 2$ .

y = 1 + 2(x - 4)

Point (4, 1), slope 2.

y = 2x - 7

Expanded: 2x − 8 + 1.

For $f(x) = x^2$ , $f'(x) = 2x$ . Draw the tangent line to the graph at $x = 1$ : place two points on it.
For $f(x) = x^2$ , $f'(x) = 2x$ . Draw the tangent line to the graph at $x = -2$ : place two points on it.

$f(1) = 2$ and $f'(1) = 3$ . Write the tangent line to the graph of $f$ at $x = 1$ : enter the right-hand side of $y = \ldots$
$f(2) = -5$ and $f'(2) = 3$ . Write the tangent line to the graph of $f$ at $x = 2$ : enter the right-hand side of $y = \ldots$

For $f(x) = x^3$ , the difference quotient at $x = -3$ simplifies to $27 + -9h + h^2$ . What is the slope of the tangent line at $x = -3$ ?
For $f(x) = x^3$ , the difference quotient at $x = -2$ simplifies to $12 + -6h + h^2$ . What is the slope of the tangent line at $x = -2$ ?

Boss problem. $f(x) = x^2 - 4x$ . Find the tangent line to its graph at $x = 5$ , from the definition.

Three questions to check you have got it.

$f(1) = 2$ and $f'(1) = 3$ . Write the tangent line to the graph of $f$ at $x = 1$ : enter the right-hand side of $y = \ldots$
$f(2) = -5$ and $f'(2) = 3$ . Write the tangent line to the graph of $f$ at $x = 2$ : enter the right-hand side of $y = \ldots$
$f(1) = -4$ and $f'(1) = -3$ . Write the tangent line to the graph of $f$ at $x = 1$ : enter the right-hand side of $y = \ldots$

The tangent line

Watch out: $y = f'(a)\,x$ has the right slope but usually misses the point.

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