Equations of tangent lines by definition (1) a. Use...

Equations of tangent lines by definition (1)
a. Use definition ( 1 ) ( $p .$ I28 ) to find the slope of the line tangent to the graph of $f$ at $P$.
b. Determine an equation of the tangent line at $P$.
c. Plot the graph of $f$ and the tangent line at $P$.
$$f(x)=x^{2}-5 ; P(3,4)$$

Key Concepts

Limit Definition of the Derivative

This concept involves finding the derivative of a function at a particular point by using the limit of the difference quotient formula. It provides a precise way to compute the instantaneous rate of change of the function by considering the limit of the average rate of change as the interval approaches zero.

Derivative as the Slope of the Tangent Line

The derivative at a given point represents the slope of the tangent line to the function at that point. It captures how the function is changing at that exact location and is used to approximate the behavior of the function locally by a straight line.

Point-Slope Form of a Line

This method is used to write the equation of a line when a point on the line and the slope are known. In the context of tangents, once the derivative (slope) at a specific point is determined, the point-slope form allows for constructing the equation of the tangent line.

Graphical Interpretation and Analysis

Plotting the function along with its tangent line helps visualize how the function behaves near a specific point. It emphasizes how the tangent line serves as the best linear approximation of the function at that point, illustrating both the function’s value and its instantaneous rate of change.

Transcript

00:01 Hello, today we're going to take a given function and a specific point, and we're going to find the slope of the tangent line of the function at that point.

00:10 We're going to determine the equation of the tangent line at that point, and then we're going to plot the function as well as the tangent line at that given point.

00:18 So first, we're going to take the derivative of the equation, or the slope, because the slope equals the first order derivative.

00:29 So for this given function of x squared, sorry about that.

00:36 So for this given function of x squared minus 5, the derivative is 2x times the derivative of x, which is 1.

00:44 So it's just 2x.

00:48 Then to get the slope at a given point, we take the x value because this is x and y, and we plug in the x value.

00:58 So f prime, we'll put that back in black, f prime of 3 equals 2 times 3 which equals 6.

01:10 So that's the slope of the tangent line at the given point.

01:18 Next we're going to determine the equation of the tangent line.

01:22 So how do we do that? well, the equation of the slope is also written as y2 minus y1 over x2 minus x1.

01:34 So to get another equation or another equation or another form of this equation, we can just multiply the denominator by both sides, and that gives us y2 minus y1 equals slope times x2 minus x1.

01:53 And another way to write this could be the function minus some initial y point equals the slope of the line times the domain minus some domain some initial domain value.

02:15 And we actually have all these values from our given point and our slope of the line at the given point.

02:22 So we have our x value, our y value, and our slope.

02:27 So we can plug those in.

02:31 So y minus 4 equals 6 times x minus 3, plugging in all those values in green.

02:42 So that gives us 6x minus 18 equals y minus 4 add 4 in both sides, and that gives us y equals 6x minus 14.

03:02 And then that's the answer to the next part.

03:05 So we have a, solution a, and solution b...

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