Can you clarify your understanding for parts (a) and (b)?

In summary, the slope of the secant line for f(x) = x^3 is 19 for part (a) and 1/(x^2 + 2x + 4) for part (b), where x does not equal 2. The slope is found using the formula m = delta(y)/delta(x) and taking into account the difference in coordinates for each point. The slope for part (b) can also be written in terms of the function f(x) = x^3, where f(x) is substituted for y in the slope formula.
  • #1
nycmathguy
Homework Statement
Find slope of the secant line.
Relevant Equations
f(x) = x^3
Slope of a Tangent Line For f (x) = x^3.

(a) Find the slope of the secant line containing the points (2, 8) and (3, 27).

(b) Find the slope of the secant line containing the points (2, 8) and (x, f (x)), where x does not equal 2.

For (a), I just have to find m, the slope using m = delta (x)/delta(y). You say?

Can someone explain part (b)?
We are given the point (x, f(x)).
I know that y = f(x). We are given f(x) = x^3.

From this piece of information, I developed the point (x, x^3). What's next?

If my understanding for parts (a) and (b) is wrong, I need someone to put my feet in the right direction. DO NOT SOLVE the problem in my stead.

Thanks
 
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  • #2
These are the ideas:

1623855402675.png

The slope of the secant is the ratio ##\dfrac{\Delta y}{\Delta x}## where ##\Delta## stands for a difference. Say point one has the coordinates ##(x_1,y_1)## and point two has the coordinates ##(x_2,y_2).## Then ##\Delta x= x_2 - x_1## and ##\Delta y= y_2-y_1.## The order is important: either second minus first coordinate on both, or the other way around on both.

The tangent is a secant where both points of intersection come closer and closer until there is only one touching point left. This means we make ##\Delta x## smaller and smaller, such that the slope of the tangent becomes ##\displaystyle{\lim_{x_1 \to x_2}\dfrac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}}.##
 
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  • #3
nycmathguy said:
Homework Statement:: Find slope of the secant line.
Relevant Equations:: f(x) = x^3
The more relevant equation would be the formula for the slope of a line segment between two points. Your relevant equation is really part of the Homework Statement.
nycmathguy said:
Slope of a Tangent Line For f (x) = x^3.

(a) Find the slope of the secant line containing the points (2, 8) and (3, 27).

(b) Find the slope of the secant line containing the points (2, 8) and (x, f (x)), where x does not equal 2.

For (a), I just have to find m, the slope using m = delta (x)/delta(y). You say?

Can someone explain part (b)?
We are given the point (x, f(x)).
I know that y = f(x). We are given f(x) = x^3.

From this piece of information, I developed the point (x, x^3). What's next?

If my understanding for parts (a) and (b) is wrong, I need someone to put my feet in the right direction. DO NOT SOLVE the problem in my stead.

Thanks
 
  • #4
nycmathguy said:
Homework Statement:: Find slope of the secant line.
Relevant Equations:: f(x) = x^3

Slope of a Tangent Line For f (x) = x^3.

(a) Find the slope of the secant line containing the points (2, 8) and (3, 27).

(b) Find the slope of the secant line containing the points (2, 8) and (x, f (x)), where x does not equal 2.

For (a), I just have to find m, the slope using m = delta (x)/delta(y). You say?
I say ...
The slope, m, of a line which passes through points (x1, y1), (x2, y2) is m = (y2 - y1) / (x2 - x1) .

In short, m = (Δy)/(Δx) . You have that inverted.

Finish part a - showing your work - before diving into part b.
Can someone explain part (b)?
We are given the point (x, f(x)).
I know that y = f(x). We are given f(x) = x^3.

From this piece of information, I developed the point (x, x^3). What's next?
 
  • #5
SammyS said:
I say ...
The slope, m, of a line which passes through points (x1, y1), (x2, y2) is m = (y2 - y1) / (x2 - x1) .

In short, m = (Δy)/(Δx) . You have that inverted.

Finish part a - showing your work - before diving into part b.

Obviously, I meant to say delta (y)/delta (x).

Part (a)

Let m = slope

Our two points: (2, 8) and (3, 27).

m = (27 - 8)/(3 - 1)

m = 19/1

m = 19

How is part (b) done?
 
  • #6
fresh_42 said:
These are the ideas:

View attachment 284551
The slope of the secant is the ratio ##\dfrac{\Delta y}{\Delta x}## where ##\Delta## stands for a difference. Say point one has the coordinates ##(x_1,y_1)## and point two has the coordinates ##(x_2,y_2).## Then ##\Delta x= x_2 - x_1## and ##\Delta y= y_2-y_1.## The order is important: either second minus first coordinate on both, or the other way around on both.

The tangent is a secant where both points of intersection come closer and closer until there is only one touching point left. This means we make ##\Delta x## smaller and smaller, such that the slope of the tangent becomes ##\displaystyle{\lim_{x_1 \to x_2}\dfrac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}}.##
Part (a)

Let m = slope

Our two points: (2, 8) and (3, 27).

m = (27 - 8)/(3 - 1)

m = 19/1

m = 19

How is part (b) done?
 
Last edited by a moderator:
  • #7
nycmathguy said:
Obviously, I meant to say delta (y)/delta (x).

Part (a)

Let m = slope

Our two points: (2, 8) and (3, 27).

m = (27 - 8)/(3 - 1)
Typo. It is ##m=\dfrac{27-8}{3-2}=19.##
nycmathguy said:
m = 19/1

m = 19

How is part (b) done?
The exact same way. It doesn't matter which way a point is actually described. You simply use ##(x,y)=(x,f(x))## instead of the specific numbers ##(3,27)##.

Imagine (draw) a straight line through the points ##(10\,\text{min}\, , \,5\,\text{miles})\; , \;(30\,\text{min}\, , \,15\,\text{miles})## and calculate the slope. What does it tell you (part (a)) and what do you get if you consider the points ##(0\,\text{min}\, , \,0\,\text{miles}) ## and ##(x\,\text{min}\, , \,f(x)\,\text{miles})## instead (part (b))?
 
  • #8
fresh_42 said:
Typo. It is ##m=\dfrac{27-8}{3-2}=19.##

The exact same way. It doesn't matter which way a point is actually described. You simply use ##(x,y)=(x,f(x))## instead of the specific numbers ##(3,27)##.

Imagine (draw) a straight line through the points ##(10\,\text{min}\, , \,5\,\text{miles})\; , \;(30\,\text{min}\, , \,15\,\text{miles})## and calculate the slope. What does it tell you (part (a)) and what do you get if you consider the points ##(0\,\text{min}\, , \,0\,\text{miles}) ## and ##(x\,\text{min}\, , \,f(x)\,\text{miles})## instead (part (b))?

Let m = slope of the secant line.

Our points: (2, 8) and (x, f (x))

I can let f(x) = x^3.

m = (x^3 - 8)/(x - 2)

On top we have the difference of cubes.

m = (x - 2)/(x - 2)(x^2 + 2x + 4)

m = 1/(x^2 + 2x + 4)

Yes?
 
  • #9
nycmathguy said:
Let m = slope of the secant line.

Our points: (2, 8) and (x, f (x))

I can let f(x) = x^3.

m = (x^3 - 8)/(x - 2)

On top we have the difference of cubes.

m = (x - 2)/(x - 2)(x^2 + 2x + 4)

m = 1/(x^2 + 2x + 4)

Yes?
Yes, that is the general slope, which varies if we consider different points. We are allowed to divide by ##x-2## because ##x\neq 2## has explicitly be mentioned.

You should try my simpler example with the car to see why the slope is an interesting quantity.
 
  • #10
nycmathguy said:
m = (x^3 - 8)/(x - 2)
On top we have the difference of cubes.
Right.

nycmathguy said:
m = (x - 2)/(x - 2)(x^2 + 2x + 4)
m = 1/(x^2 + 2x + 4)
No, you have switched numerator and denominator.

fresh_42 said:
Yes, that is the general slope
I don't think you read what nycmathguy wrote very carefully...

Also, (x - 2)/(x - 2)(x^2 + 2x + 4) is ambiguous.
Taken literally, this is $$\frac{x - 2}{x - 2} (x^2 + 2x + 4)$$

What you meant probably was $$\frac{x - 2}{(x - 2)(x^2 + 2x + 4)}$$

Writing the expression like this -- (x - 2)/[(x - 2)(x^2 + 2x + 4)] -- would have eliminated the ambiguity, but is incorrect for the reason mentioned above.
 
  • #11
Mark44 said:
I don't think you read what nycmathguy wrote very carefully...
Yes, I was so focused on the polynomials that I didn't recognize that he switched numerator and denominator. The linear notation didn't help either.

@nycmathguy:
Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/
 
  • #12
fresh_42 said:
Yes, I was so focused on the polynomials that I didn't recognize that he switched numerator and denominator. The linear notation didn't help either.

@nycmathguy:
Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/
I appreciate the LaTex link but I'd rather use the MathMagic Lite app. A million times easier to use and learn than LaTex.
 

Related to Can you clarify your understanding for parts (a) and (b)?

1. What is the definition of slope of a secant line?

The slope of a secant line is the measure of the steepness of a line connecting two points on a curve. It is calculated by dividing the change in the y-coordinates by the change in the x-coordinates between the two points.

2. How do you find the slope of a secant line?

To find the slope of a secant line, you need to choose two points on the curve and calculate the change in the y-coordinates and the change in the x-coordinates between those points. Then, divide the change in y by the change in x to get the slope.

3. What is the difference between slope of a secant line and slope of a tangent line?

The slope of a secant line measures the average rate of change between two points on a curve, while the slope of a tangent line measures the instantaneous rate of change at a specific point on the curve.

4. How is the slope of a secant line used in calculus?

The slope of a secant line is used in calculus to approximate the slope of a tangent line at a specific point on a curve. As the two points on the secant line get closer and closer together, the slope of the secant line approaches the slope of the tangent line.

5. Can the slope of a secant line be negative?

Yes, the slope of a secant line can be negative. This means that the curve is decreasing between the two points where the secant line intersects it. A negative slope indicates that the curve is sloping downwards from left to right.

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