Question

how would you find an equation of a tangent line to a curve at a point when the point is not on the curve?

Answer 1

Does this make sense to anybody else?

Answer 2

there is no tangent line to a point

Answer 3

you can not find that.. it doesnt make sense

Answer 4

cant find such line

Answer 5

do you have such a homework problem?

Answer 6

you actually can, first you have to find the derivative of the function of the curve given n then i am confused what to do, yes it was a homework problem, but it is such a difficult one that my teacher made it an extra credit problem!

Answer 7

what your asking for and what youve typed are 2 different things i believe

Answer 8

no it makes no sense....ur asking how you would find a tangent line to a curve at a certain point on that curve, but its not on the curve. that's contradictory

Answer 9

you might be wanting to find the tangent to the curve that includes the given point that is not on the line

Answer 10

i think i am confusing everybody, rather i should type the problem itself

Answer 11

lol .... that would be a good start yes

Answer 12

=)

Answer 13

here is the problem: Find equations of the tangent lines to the graph of f(x)= x/x-1 that passes through the point (-1,5)

Answer 14

yay!! i was right

Answer 15

youneed it in point slope form or slope intercept form?

Answer 16

that point is not on the curve given

Answer 17

yea....

Answer 18

i know thats what makes this problem hard!!!! its not on the curve

Answer 19

its not on the given curve; but there is a tangent to the curve that will go thru it

Answer 20

maybe a few

Answer 21

yes, if you graph it using graphing calculator, u will see there r 2 possible places there can be tangent lines to the curve n that can pass through the point, but u need to find the equation of either of those tangent lines

Answer 22

(x,f(x))
-(-1, 5)
--------
(x+1,f(x)-5); slope = \(\cfrac{f(x)-5}{x+1}\)

this is the slope of the line between the function and the point

Answer 23

now when does the derivative of the function equal this slope?

Answer 24

this is the function right?
\[f(x)=\frac{x}{x-1}\]

Answer 25

well i think ur on the right track with the slope, but im not sure if the slope n the derivative has to be the same or not!!

Answer 26

yes that is the function!!

Answer 27

oh i thought he was saying he wanted the tangent line of the curve where the curve passes through (-1,5)
he wants a tangent line to some point on the curve that passes through (-1,5)

Answer 28

\[f(x)=\frac{x}{x-1}\]
\[f'(x)=\frac{(x-1)x'-(x-1)'x}{(x-1)^2}\]
\[f'(x)=\frac{(x-1)-x}{(x-1)^2}\]
\[f'(x)=\frac{-1}{(x-1)^2}\]
so far if i did it right

Answer 29

\[dy/dx=-1/(x-1)^2\]. Using\[ y=mx+b\], we have \[5=1/4+b\], so b=19/4

Answer 30

i have seen this exact problem before.

Answer 31

\[y=-x/(x-1)^2+19/4\]

Answer 32

\[\cfrac{f(x)-5}{x+1}=\frac{-1}{(x-1)^2}\]
\[\cfrac{\frac{x}{x-1}-5}{x+1}=\frac{-1}{(x-1)^2}\]
\[\frac{x-5(x-1)}{(x+1)(x-1)}=\frac{-1}{(x-1)^2}\]
\[\frac{x-5x+5}{x^2-1}=\frac{-1}{x^2-2x+1}\]
the rest is just algebra i believe

Answer 33

put

\[\frac{\frac{x}{x-1}-5}{x+1}=\frac{1}{(x-1)^2}\] and solve for x i think.

Answer 34

yep, thats what id do; but i tend to do it the hardest way possible lol

Answer 35

and i am wrong in any case

Answer 36

\[\frac{\frac{x}{x-1}-5}{x+1}=-\frac{1}{(x-1)^2}\] might be more like it

Answer 37

yes looks good satellite

Answer 38

if it was me i would "cross multiply"

Answer 39

\[\frac{x-5x+5}{x^2-1}=\frac{-1}{x^2-2x+1}\]
\[(-4x+5)(x^2-2x+1)=-x^2+1\]
\[-4x^3+13x^2-14x+5=-x^2+1\]
\[-4x^3+13x^2+x^2-14x+5-1=0\]
\[-4x^3+14x^2-14x+4=0\]

Answer 40

might be able to group that into a factor

Answer 41

Let me get the problem straight: we are trying to figure out the equation of the tangent line that passes through the point (-1,5), rite?

Answer 42

\[(x-1) \cdot \frac{\frac{x}{x-1}-5}{x+1}=-\frac{1}{(x-1)^2} \cdot (x-1)\]
\[\frac{x-5(x-1)}{x+1}=-\frac{1}{x-1}\]
\[\frac{-4x+5}{x+1}=\frac{-1}{x-1} => (-4x+5)(x-1)=-(x+1)\]
\[-4x^2+4x+5x-5+x+1=0\]
\[-4x^2+10x-4=0 =>2x^2-5x+2=0\]

Answer 43

\[-4x^3+14x^2-14x+4 =0\]

just let the wolf do it ...

http://www.wolframalpha.com/input/?i=-4x^3%2B14x^2-14x%2B4+%3D0

we get 3 roots to try out

Answer 44

yes, that is how i see it

Answer 45

\[2x^2-4x-x+2=0 => 2x(x-2)-1(x-2)=0\]
\[(x-2)(2x-1)=0 => x=2 \text{ or } x=\frac{1}{2}\]

Answer 46

i got 1/2 1 or 2

Answer 47

1 is not in the domain

Answer 48

\[\frac{-4.1+5}{1^2-1}=\frac{-1}{1^2-2.1+1}\]
\[\frac{-4.1+5}{1-1}=\frac{-1}{1^2-2.1+1};\ bad\]
\[\frac{-4.2+5}{2^2-1}=\frac{-1}{2^2-2.2+1}\]
\[\frac{-8+5}{4-1}=\frac{-1}{4-4+1}\]
\[\frac{-3}{3}=-1\; good\]
\[\frac{-4x+5}{x^2-1}=\frac{-1}{x^2-2x+1}\]
\[\frac{-4..5+5}{.5^2-1}=\frac{-1}{.5^2-2..5+1}\]
\[\frac{-2+5}{.25-1}=\frac{-1}{.25-1+1}\]
\[\frac{3}{-.75}=\frac{-1}{.25};good\]

Answer 49

i actually found what to do, i m not sure if u've done it right or not

Answer 50

after finding the derivative, u do slope intercept form, use the derivative as ur slope n use the point given too, then simplify

Answer 51

using the point given in the derivative only works if the given point is ON the curve

Answer 52

nope, i know this graph looks like this

Answer 53

so what's the answer you're getting?

Answer 54

take a look at this example, u'll get a clear picture---
Here's a slightly more complicated variant on the same problem. Find the two lines that are tangent to

y = x2 - 2x + 1
and pass through the point, (5, 7). Observe that in this case, the point, (5, 7), does not lie on the curve. So you are finding lines that pass through a point outside the curve but at some other points, which are as yet unknown, they are tangent to the curve. The strategy is to identify those points of tangency. From them, it is easy to find the lines that solve the problem.
Step 1: Find the derivative of the curve. In this case we have

y' = 2x - 2
We do this because we know that at the point of tangency, the derivative of the curve must be equal to the slope of the tangent line. So we need to know this derivative.
Step 2: Write as much of the equation of the line as you can from what you know. The slope, m, of the line is still unknown. But you know what point it must pass through. In this case that is the point (5,7). So using the formula for the equation of a line passing through a given point with a given slope, you have:

y - 7 = m(x - 5)
Step 3: Use the derivative to write an equation for the slope. We don't know the slope of the line yet, but we know that if the point, (x,y), is the point of tangency, then the slope of the line will be equal to the derivative of the original curve at x. The derivative we already determined to be y' = 2x - 2. So we write the equation

m = y' = 2x - 2
Step 4: Substitute the slope expression into the equation for the line. You have m = 2x - 2 and y - 7 = m(x - 5). Put them both together by substituting for m in the second equation.

y - 7 = (2x - 2)(x - 5) = 2x2 - 12x + 10

y = 2x2 - 12x + 17
Step 5: Substitute the equation for the original curve back in. That is, you know that y = x2 - 2x + 1, where (x,y) is the point of tangency. Why? Because the point of tangency must lie on the curve. Substituting that expression for y into the above gives:

x2 - 2x + 1 = 2x2 - 12x + 17

Step 6: Gather like terms and solve for x. Using simple algebra the above equation becomes the quadratic

x2 - 10x + 16 = 0
You can either use the quadratic formula, or you can factor this one in your head to get
(x - 2)(x - 8) = 0
So the x coordinate of the the point of tangency is either x = 2 or x = 8.
Step 7: Substitute back to get m. That is, we have already seen that m = 2x - 2 at any point, (x, y), of tangency. Since we now know what both the x's are for the points of tangency, we can put those values into the equation for m and get that either m = 2 or m = 14. From the equation we had in step 2,

y - 7 = 2(x - 5)
and
y - 7 = 14(x - 5)
When you multiply those out and put them into slope-intercept form, the equations of the two lines that are tangent to y = x2 - 2x + 1 and pass through the point, (5, 7), are
y = 2x - 3
and
y = 14x - 63
and you are done. The curve and the two lines are illustrated in the graph on the right here. Notice where the two lines intersect. Both points of tangency are visible on the graph, although one is very nearly at the top of the graph.

Answer 55

that is one method yes.

Answer 56

its actually the easiest one i believe

Answer 57

lets try it then

\[\frac{x}{x-1}=\frac{-x}{(x-1)^2}-\frac{1}{(x-1)^2}+5\]
\[\frac{x(x-1)}{(x-1)^2}=\frac{-x-1+5(x-1)^2}{(x-1)^2}\]
\[x(x-1)=-x-1+5(x-1)^2\]
\[x^2-x=-x-1+5(x^2-2x+1)\]
\[x^2=-1+5x^2-10x+5\]
\[x^2=5x^2-10x+4\]
\[0=4x^2-10x+4\]
\[x=\frac{10\pm\sqrt{100-4(16)}}{8}\]
\[x=\frac{10\pm\sqrt{36}}{8}\]
\[x=\frac{10\pm6}{8};2,1/2\]
seems to work

Answer 58

YES is does! nice job =)

Answer 59

equations are y= -x+4 and y=-4x+1