Secant Lines Consider the function f(x)=6 x-x^2 and the...

Secant Lines Consider the function $f(x)=6 x-x^{2}$ and the point $P(2,8)$ on the graph of $f$ . (a) Graph $f$ and the secant lines passing through $P(2,8)$ and $Q(x, f(x))$ for $x$ -values of $3,2.5,$ and $1.5 .$ (b) Find the slope of each secant line. (c) Use the results of part (b) to estimate the slope of the tangent line to the graph of $f$ at $P(2,8) .$ Describe how to improve your approximation of the slope.

Secant Lines Consider the function $f(x)=6 x-x^{2}$ and the point $P(2,8)$ on the graph of $f$ .
(a) Graph $f$ and the secant lines passing through $P(2,8)$ and $Q(x, f(x))$ for $x$ -values of $3,2.5,$ and $1.5 .$
(b) Find the slope of each secant line.
(c) Use the results of part (b) to estimate the slope of the tangent line to the graph of $f$ at $P(2,8) .$ Describe how to improve your approximation of the slope.

Key Concepts

Derivative as Limit of Secant Slopes

The derivative of a function at a point is defined as the limit of the slopes of secant lines as the second point approaches the point of tangency. This limit process transforms the average rate of change into an instantaneous rate of change, thereby providing the slope of the tangent line to the curve at that point.

Secant Line

A secant line is a straight line that intersects a curve at two or more points. In a calculus context, it is used to average the rate of change of a function between two points on its graph, providing a finite difference approximation of the function's behavior over that interval.

Slope Calculation

The slope of a line, including a secant line, is calculated as the ratio of the change in the y-values to the change in the x-values between two points. In problems involving functions, this is often expressed as [f(x2) - f(x1)] / [x2 - x1]. This calculation gives an average rate of change for the function over the interval between the two points.

Tangent Line

A tangent line to a curve at a specific point is a line that touches the curve at that point without cutting through it. It represents the instantaneous direction of the curve, making it the best linear approximation of the function near that point.

Transcript

00:01 In problem 7 we have a function of x equals 6 x minus x squared and a point on f which has a coordinates 2 and 8 part a we want to graph f and the second lines let's first graph if to graph f we'll make here x and y 1 2 3 4 5 we have here 1 2 3 4 5 6 7 8 and we have minus 1 minus 2 minus 3 minus 4 let's substitute by x equal 0 we have f equals 0 when x equals 1 we have f equals 6 minus 1 equals 5 and when we have x equals minus 1 and when we have x equals minus 1 we have minus 6 minus 1 equals minus minus 6 minus 1 equals minus 7 1 2 3 4 5 6 7 then we have a point here and when x equals 2 we have y equals 8 when x equals 3 we have y equals 3 multiplied by 6 minus 9 equals 9 here and when x equals 4 we have 6 multiple 4 minus 4 squared equals 8 and we can see that at x equals 5 we have y equals 5 and so on this is a graph we can make it more accurate by a free hand we can sketch this graph like that we have the point b here we want also to sketch the second lines passing through the point p 2 and 8 and the point on the graph which has the value of x it's x coordinates 3 2 .5 and 1 .5 is here 1 .5 is here then we can determine this point on the graph it will be here let's change the color we have at 1 .5 5 we have this point here at 1 .5 and at 2 .5 we have this point here and at 3 we have this point here.

03:41 The second lines the first second line is this line the second one is this line and finally the third line is this line.

04:01 For bar to be we want to find the slope of each second line we can calculate the slope of the first second line this is the first one as slope in general equals delta y divided by delta x then slope of the first second line that connects the point b2 and 8 and this point this point has a coordinate 1 .5 and we can get its y coordinate by substituting in the function by x equals 1 .5 it equals 9 minus 1 .5 squared equals 6 .75 then the slope equals 8 minus 6 .75 divided by 2 minus 1 .5 which equals 2 .5 this is for the first line here the second line it connects the point 2 and 8 and the point which has the x coordinate of 2 .5 we can get the y coordinate by substituting in the function here 6 multiplied by 2 .5 minus 2 .5 squared equals 8 .75 then the slope for the second line which is this line equals 8 minus 8 .75 divided by 2 minus 2 .5...

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