# Math Help - Finding the tangent and secants of a point of a given function?

1. ## Finding the tangent and secants of a point of a given function?

I am really struggling in understanding this concept (mainly because I'm doing my math course online, and it's pre-calculus and not calculus), and I couldn't find any thing on this concept at all online after multiple tries on different search engines. I know it has something to do with instantaneous velocity (instantaneous rate of change), but my book doesn't explain any further than that.

If anyone can point me in the direction(helpful links) or explain the concepts to me, I'd greatly appreciate it. Thank you in advance.

2. Post a particular question you are having problems with.

3. Alright, here's the question:
Answer the following questions, given the equation $-3(1/2)^x^-^2+6$

a) Graph the function using a suitable scale and any means at your disposal (this I can do, as I have graphing software).

b) Draw the secants from y-intercept to each of the points on the curve ending at $x=2,1,0.5,0.1,0.01,0.001$ (this confuses me, because there were no x and y given for the equation which I would have made equal to x2 and y2).

c) Determine the equation of each of the secants constructed in part b (I usually do this part before graphing the secants in order to make it easier to do the points).

d) determine the equation of the tangent at y-intercept in simplified form (I have absolutely no idea on how to answer this one).

I would greatly appreciate it if you could explain all the steps you took, I kind of understand the concepts so far but not in clarity (I am thinking x = 0, because the secant points are approaching the zero, but I could be wrong).

4. Originally Posted by Pupil
b) Draw the secants from y-intercept to each of the points on the curve ending at $x=2,1,0.5,0.1,0.01,0.001$ (this confuses me, because there were no x and y given for the equation which I would have made equal to x2 and y2).

c) Determine the equation of each of the secants constructed in part b (I usually do this part before graphing the secants in order to make it easier to do the points).
Lets start with these two parts. With these points $x=2,1,0.5,0.1,0.01,0.001$ one at a time compare them to the y-intercept (x=0) like this

x=0: gives $-3(1/2)^{(0)-2}+6=\dots$ what do you get?

x=2: gives $-3(1/2)^{(2)-2}+6=\dots$ what do you get?

x=1: gives $-3(1/2)^{(1)-2}+6=\dots$ what do you get?

x=0.5: gives $-3(1/2)^{(0.5)-2}+6=\dots$ what do you get?

and so on....

then draw the lines connecting

$(0, f(0))$ and $(2, f(2))$

$(0, f(0))$ and $(1, f(1))$

$(0, f(0))$ and $(0.5, f(0.5))$

and so on...

Then find the equation of the straight line between each of these pairs.

What do you notice about the gradient of these lines as $x$ gets very small?

5. Originally Posted by Pupil
d) determine the equation of the tangent at y-intercept in simplified form (I have absolutely no idea on how to answer this one).
Ever heard of differentiation?

I find the Khan academy videos for calculus are very educational.

The y-intercept is when x = 0. To find the equation of the tangent line you would need to find the gradient by differentiation and then find the C constant by using the initial equation.

6. Originally Posted by pickslides
Lets start with these two parts. With these points $x=2,1,0.5,0.1,0.01,0.001$ one at a time compare them to the y-intercept (x=0) like this

x=0: gives $-3(1/2)^{(0)-2}+6=\dots$ what do you get?

x=2: gives $-3(1/2)^{(2)-2}+6=\dots$ what do you get?

x=1: gives $-3(1/2)^{(1)-2}+6=\dots$ what do you get?

x=0.5: gives $-3(1/2)^{(0.5)-2}+6=\dots$ what do you get?

and so on....

then draw the lines connecting

$(0, f(0))$ and $(2, f(2))$

$(0, f(0))$ and $(1, f(1))$

$(0, f(0))$ and $(0.5, f(0.5))$

and so on...

Then find the equation of the straight line between each of these pairs.

What do you notice about the gradient of these lines as $x$ gets very small?
At $x=0$I get $f(0)=-6$ Which I've come to find is the original function.

At $x=2$ I get $f(2)=3$ So this would be my first secant, correct? $x2=0, x1=2, y2=-6, y1=3$? And then rinse and repeat for the rest of the secants, connecting them with the original point (0,-6)? So the equation for the first secant would be $4.5x-6$ right?

I notice that, while $x$ may be gradually decreasing in value as it approaches (0,-6) but the slope increases in value.

7. Originally Posted by Educated
Ever heard of differentiation?

I find the Khan academy videos for calculus are very educational.

The y-intercept is when x = 0. To find the equation of the tangent line you would need to find the gradient by differentiation and then find the C constant by using the initial equation.
I watched the video and it seems too calculus heavy on me for now, he applied concepts and terms that are foreign to me. Do I need a solid understanding of calculus in order to get differentiation, or could I learn it now without knowing anything about calculus yet?

8. Originally Posted by Educated
Ever heard of differentiation?