# Thread: Equation of Secant Line

1. ## Equation of Secant Line

Given f(x) = -3x + 2, find the equation of the secant line at x = 1 with h = 0.01.

The book instructs students to use

M_sec = [f(x + h) - f(x)]/h

I have seen this expression before. This is the difference quotient of f, is it not?

I am seeking the needed steps that will guide me through this problem.

2. ## Re: Equation of Secant Line

Did you do any "googling"?

3. ## Re: Equation of Secant Line

Originally Posted by harpazo
Given f(x) = -3x + 2, find the equation of the secant line at x = 1 with h = 0.01.

The book instructs students to use

M_sec = [f(x + h) - f(x)]/h

I have seen this expression before. This is the difference quotient of f, is it not? I haven't heard it called that, but it is simply the gradient of the line required.

I am seeking the needed steps that will guide me through this problem.
The secant line joins two points on the curve. One point is (1, f(1)) and the other is (1.01, f(1.01)).

You now need to find the equation of the line through these two points n the usual way.

4. ## Re: Equation of Secant Line

Just going by intuition first, what do you suspect the secant line to a linear function is going to be?

5. ## Re: Equation of Secant Line

Originally Posted by MarkFL
Just going by intuition first, what do you suspect the secant line to a linear function is going to be?
Oh, of course, didn't take any notice of f(x).

6. ## Re: Equation of Secant Line

Originally Posted by DenisB
Did you do any "googling"?
No.

7. ## Re: Equation of Secant Line

Originally Posted by MarkFL
Just going by intuition first, what do you suspect the secant line to a linear function is going to be?
Perpendicular.

8. ## Re: Equation of Secant Line

Originally Posted by harpazo
Perpendicular.
No.

Do you know what the secant of a function looks like? (Hint: Think of the secant of a circle.)

I don't mean to criticize but I really don't think you are ready for this kind of problem. This is Pre-Calculus and you've had a lot of problems with Algebra. I'd advise more Algebra review.

-Dan

9. ## Re: Equation of Secant Line

Originally Posted by topsquark
No.

Do you know what the secant of a function looks like? (Hint: Think of the secant of a circle.)

I don't mean to criticize but I really don't think you are ready for this kind of problem. This is Pre-Calculus and you've had a lot of problems with Algebra. I'd advise more Algebra review.

-Dan
This problem is from my Michael Sullivan College Algebra 9th Edition textbook. I think I can do it. Thanks anyway.

10. ## Re: Equation of Secant Line

Let's look at the definition of a secant line:

In geometry, a secant of a curve is a line that intersects the curve in at least two points. The word secant comes from the Latin word secare, meaning to cut.

So, imagine a line in the plane, and then further imagine two points on that line and a line though those two points. Can you see that the secant line will coincide with the original line?

Consider the linear function:

$\displaystyle f(x)=ax+b$

Now, consider two arbitrary points on this linear function:

$\displaystyle (x,f(x))=(x,ax+b)$

$\displaystyle (x+h,f(x+h))=(x+h,a(x+h)+b)$

And so the slope $\displaystyle m$ of the secant line is given by:

$\displaystyle m=\frac{f(x+h)-f(x)}{(x+h)-(x)}=\frac{(ax+ah+b)-(ax+b)}{h}=a$

Then using the point slope formula, we find the equation of the secant line to be:

$\displaystyle y=a(x-x)+f(x)=f(x)$

Thus, we find the secant line does in fact coincide with the original line.

11. ## Re: Equation of Secant Line

Originally Posted by MarkFL
Let's look at the definition of a secant line:

In geometry, a secant of a curve is a line that intersects the curve in at least two points. The word secant comes from the Latin word secare, meaning to cut.

So, imagine a line in the plane, and then further imagine two points on that line and a line though those two points. Can you see that the secant line will coincide with the original line?

Consider the linear function:

$\displaystyle f(x)=ax+b$

Now, consider two arbitrary points on this linear function:

$\displaystyle (x,f(x))=(x,ax+b)$

$\displaystyle (x+h,f(x+h))=(x+h,a(x+h)+b)$

And so the slope $\displaystyle m$ of the secant line is given by:

$\displaystyle m=\frac{f(x+h)-f(x)}{(x+h)-(x)}=\frac{(ax+ah+b)-(ax+b)}{h}=a$

Then using the point slope formula, we find the equation of the secant line to be:

$\displaystyle y=a(x-x)+f(x)=f(x)$

Thus, we find the secant line does in fact coincide with the original line.
Nicely-done!

12. ## Re: Equation of Secant Line

Originally Posted by harpazo
This problem is from my Michael Sullivan College Algebra 9th Edition textbook. I think I can do it. Thanks anyway.
No offense intended, but I do believe a College Algebra text is beyond you in many topics. In many respects College Algebra and Pre-Calculus overlap. Still, let's give it a try.

-Dan

13. ## Re: Equation of Secant Line

Originally Posted by Debsta
The secant line joins two points on the curve. One point is (1, f(1)) and the other is (1.01, f(1.01)).

You now need to find the equation of the line through these two points n the usual way.
Just repeating my earlier comment above.

If the function is a straight line, then the secant (the straight line joining any two points on the curve think about it!) must be the same straight line.

Doing it the long way and using the method implied in the question "Given f(x) = -3x + 2, find the equation of the secant line at x = 1 with h = 0.01":

If x=1, y= -3*1+2 = -1, so one point is (1, -1).

If x= 1.01, y = -3*1.01 + 2 = -1.03 , so other point is (1.01, -1.03)

Gradient of secant = (-1.03 +1)/(1.01 - 1) =-0.03/0.01 = -3

Using y - y1 = m(x - x1) where m = -3 and (x1, y1)=(1, -1):

y + 1 =-3 (x-1)

y = -3x +3 -1

y=-3x +2 (confirming that the secant is the same line (obviously))

This is the method to use if the original function is not a straight line.

14. ## Re: Equation of Secant Line

Originally Posted by Debsta
Just repeating my earlier comment above.

If the function is a straight line, then the secant (the straight line joining any two points on the curve think about it!) must be the same straight line.

Doing it the long way and using the method implied in the question "Given f(x) = -3x + 2, find the equation of the secant line at x = 1 with h = 0.01":

If x=1, y= -3*1+2 = -1, so one point is (1, -1).

If x= 1.01, y = -3*1.01 + 2 = -1.03 , so other point is (1.01, -1.03)

Gradient of secant = (-1.03 +1)/(1.01 - 1) =-0.03/0.01 = -3

Using y - y1 = m(x - x1) where m = -3 and (x1, y1)=(1, -1):

y + 1 =-3 (x-1)

y = -3x +3 -1

y=-3x +2 (confirming that the secant is the same line (obviously))

This is the method to use if the original function is not a straight line.
Thank you for your kindness and patience.