Find the slope of the tangent line to the graph of the function at the given point, and determin

step 1 We're looking at a tangent line problem. And with a tangent line, what we need is a point and a

Find the slope of the tangent line to the graph of the...

Key Concepts

Derivative

A derivative represents the rate at which a function's value changes with respect to changes in its input. In the context of tangent lines, the derivative evaluated at a particular point gives the slope of the tangent line at that point.

Tangent Line

A tangent line to a curve at a specific point is a straight line that touches the curve exactly at that point and has a slope equal to the derivative of the function at that point. It provides the best linear approximation of the function near that point.

Point-Slope Form

The point-slope form of a line is an algebraic expression used to define a line given a point and its slope. It is typically written as y - y? = m(x - x?), where m is the slope and (x?, y?) is a point on the line. This form is essential for writing the equation of a tangent line when the slope and a point on the curve are known.

Differentiation Rules

Differentiation rules, such as the power rule, product rule, quotient rule, and chain rule, are techniques used to calculate the derivative of a function. These rules simplify the process of finding the slope of the tangent line by providing systematic methods for handling various types of functions.

Transcript

00:01 We're looking at a tangent line problem.

00:03 And with a tangent line, what we need is a point and a slope.

00:09 And they generously tell us the point is at negative 2, negative 10.

00:16 And in order to find the slope, what we need to do is find the derivative and plug in the x coordinate negative 2 into the derivative.

00:26 And they also tell us that f of x is equal to 3x minus x squared.

00:33 So the first thing that we need to do, since we're doing the limit definition of the directive, is figure out what f of x plus h is.

00:44 So i need to replace all of those x's in the equation with x plus h.

00:50 And when i start to distribute as well as folio, because remember x plus h squared means x plus h times x plus h.

01:01 And while i'm added, i also need to distribute that negative into that.

01:07 So there would be 2xh and then also an h squared that would become negative.

01:13 So when i go to find the limit as h approaches zero of f of x plus h minus the difference of those two, what you'll notice is the 3x minus 3x will cancel, and then the negative x squared minus a negative x squared.

01:30 Will cancel.

01:32 And what i can do is factor out an h between the 3, the minus 2x, and the minus h.

01:39 And that way, i could cancel out the h's.

01:43 By the way, this is the definition of f prime of x is what i'm doing.

01:47 And now i can plug in zero for this, for the h value, i'm left with 3 minus 2x...

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