Thread: Difference quotient / secant line

1. Difference quotient / secant line

I dont understand how the difference quotient of f works.... the problem is

f(x) = 3x^2 + x

A.) express the slope of the secant line of each function in terms of x and h.

B.) Find the equation for the secant line at x = 1 with h = 0.01

Any help would be appreciated thank you

2. Originally Posted by cokeclassic
I dont understand how the difference quotient of f works.... the problem is

f(x) = 3x^2 + x

A.) express the slope of the secant line of each function in terms of x and h.
just plug this into the formula for the difference quotient.

$\displaystyle \text{Slope} = \frac {f(x + h) - f(x)}h$

B.) Find the equation for the secant line at x = 1 with h = 0.01
plug in x = 1 and h = 0.01 above and solve.

this gives you the slope of the secant line, thus it is your m in the equation for the line, y = mx + b

the line passes through the point (1,4). (plug in the x-value into f(x) to find the corresponding y-value) you can now find the equation of the line

see here for two methods on finding the equation of a line given the slope and a point.

3. how do you
A.) express the slope of the secant line of each function in terms of x and h.

I think its suppose to be a equation to find the slope?

I was told you plug in the points after you get part A to get the slope for part B

and how do you know it passes through (1,4)?

4. Originally Posted by cokeclassic
how do you
A.) express the slope of the secant line of each function in terms of x and h.

I think its suppose to be a equation to find the slope?
i gave you the equation in my last post. just plug in the pieces.

I was told you plug in the points after you get part A to get the slope for part B
yes, that is when you are finding the equation of the line. you have to get the slope first, which is what part (a) shows you how to do, and (b) makes it specific

and how do you know it passes through (1,4)?
i also told you this in my last post. the y value corresponding to x = 1 is f(1)

5. for part A I have to plug in f(x) = 3x^2 + x into

f(x + h) - f(x) / h

but I dont understand how to do that

6. Originally Posted by cokeclassic
for part A I have to plug in f(x) = 3x^2 + x into

f(x + h) - f(x) / h

but I dont understand how to do that
do you know what composite functions are? whatever shows up in the brackets, you replace x with it

so if, say, $\displaystyle f(x) = x^2 + 2x + 3$

then $\displaystyle f(1) = (1)^2 + 2(1) + 3 = 6$

$\displaystyle f(x + 2) = (x + 2)^2 + 2(x + 2) + 3$

$\displaystyle f( \sqrt{x}) = (\sqrt{x})^2 + 2( \sqrt{x}) + 3$

$\displaystyle f($$\displaystyle ) = ($$\displaystyle )^2 + 2($$\displaystyle ) + 3$ ...........provided is in the domain of f of course

$\displaystyle f(x + h) = (x + h)^2 + 2(x + h) + 3$

now can you continue?

7. hey jhevon I am having another issue with a simular problem

problem is f(x) = x^2 + 2x
The question is the same
A.)express the slope of the secant line of each function in terms of x and h.

Answer for A is : 2x + h + 2

B.) Find the equation for the secant line at x = 1 with h = 0.01

so I plug in the numbers into 2x + h +2 and get the slope 4.01x

but how do I get the Y intercept??

the answer for B is Y = 4.01x - 1.01

I just dont know where the 1.01 came from

8. Originally Posted by cokeclassic
hey jhevon I am having another issue with a simular problem

problem is f(x) = x^2 + 2x
The question is the same
A.)express the slope of the secant line of each function in terms of x and h.

Answer for A is : 2x + h + 2

B.) Find the equation for the secant line at x = 1 with h = 0.01

so I plug in the numbers into 2x + h +2 and get the slope 4.01x

but how do I get the Y intercept??

the answer for B is Y = 4.01x - 1.01

I just dont know where the 1.01 came from
did you get the last question?

the link i gave you in my first post will tell you how they got it. they just solved for the equation of the line, which the link shows you how to do

9. I saw it but for either of those methods what do I plug in.. for example in method 1 what do i plug in for y1? and method 2 for b?

im giving the x and the h no y tho :/

i tried plugging in any 2 points that fall on the graph of 2x +h +1 but the answer just doesnt come out right

10. I think I finally got it.... so basically the h = 0.01 is useless in finding the equation? all I do get the y intercept for x^2 + 2x which is 3, and then plug those into point-slope get y-3 = 4.01(x-1) and then y = 4.01 - 1.01 which is the answer? was that correct?

11. Originally Posted by cokeclassic
I think I finally got it.... so basically the h = 0.01 is useless in finding the equation? all I do get the y intercept for x^2 + 2x which is 3, and then plug those into point-slope get y-3 = 4.01(x-1) and then y = 4.01 - 1.01 which is the answer? was that correct?
not the y-intercept (that would be 0), the y-value when x = 1. so yes, it is 3. so the secant line gives the slope m and you have the point (1,3) and you use that as your $\displaystyle (x_1,y_1)$ in the formula $\displaystyle y - y_1 = m(x - x_1)$

12. Ok thank you very much for the time