Question

There are many tangent lines to the graph of f(x) = (x-2)^1/2

Find equations of tangent lines that pass through the point (5,2). Note the point (5,2) is not on the graph of f.

I need a hint.
I know its going to be some (a,b) on the graph that has tangent line going through it s.t. it also passes thro. (5,2).

Answer 1

find the derivative first - this will give you the slope of the tangent line at any point of the graph.
then find the value of the slope when x=5.
you should hopefully be able to solve the rest yourself.

Answer 2

Wai Wait.

Answer 3

Ok this is my attempt. Please let me post it out.

Answer 4

f(x) = (x-2)^{1/2}, then the derivative of f is:
f'(x) = 1/2(x-2)^{-1/2}

Answer 5

I have one major question

Answer 6

When you have the tangent line equation, is it going to be:
y = y'*(x) + b?

I know that we have to sub in (5,2) first to get the y-int, then we have to sub into the equation (a,(a-2)^1/2), BUT I have a question, is it going to be:

y = 1/2(a-2)^1/2*[a] + (whatever the y-int is) OR
y = 1/2(a-2)^1/2 +(whatever the y-int is)

Answer 7

Please let me know if you don't understand what I am trying to say.

Answer 8

sory - I just read the bit "Note the point (5,2) is not on the graph of f. " - so my suggestion above will not work. let me ready what you have written so far...

Answer 9

I am saying that do you write in the form:
y= y'*(x) + b
OR
y= y' + b <-- its because y' also has x-term in it, so I was confused).

Answer 10

equation of a straight line is: y = mx +c
where m is the slope and c is the y-intercept

Answer 11

you have found the equation of the set of "slopes" by finding the derivative of the graph:
\[f'(x) = 1/2(x-2)^{-1/2}\]

Answer 12

you also know that the tangent line must pass thru (5,2). so:
5 = m*2 + c
giving c = 5-2m

Answer 13

so we can write the equation of the line as:
y = mx + (5-2m) = m(x-2) + 5
this gives: m = (y-5)/(x-2)
so I believe you should be able to equate this to the derivative above to get the set of lines asked for

Answer 14

i.e.\[m=f'(x)\]

Answer 15

am I making any sense to you or have I completely mis-understood the problem?

Answer 16

sorry - I got the x and y mixed up above:
you also know that the tangent line must pass thru (5,2). so:
2 = m*5 + c
giving c = 2-5m

Answer 17

Ok let me tell you in words as to what my procedure is :
1) y = mx + b, where m = y'
2) use y = y'*x + b and sub in (5,2) to find y-int.
3) now we know there is a point (a,(a-2)^1/2) on curve s.t. its tangenet line crosses thro. (5,2)
4) Sub in (a,(a-2)^1/2 into the tangent line equation and solve for a.

Answer 18

so we can write the equation of the line as:
y = mx + (2-5m) = m(x-5) + 2
this gives: m = (y-2)/(x-5)

Answer 19

yes - you are correct

Answer 20

But it gets so ugly. How am I suppose to solve for a with all those radicals around?

Answer 21

You know the equation of any tangent line: you have its slope and you know one point on the line; namely the point on the graph of f to which it is tangent.

Write down the general equation of a tangent line.

Then see whether or not it passes through (5,2)

Answer 22

I end up with:\[y=2+\frac{x-5}{2\sqrt{x-2}}\]

Answer 23

f(x) = (x-2)^1/2 , Now for a point x = a, f'(a) = (1/2)(a-2)^(-1/2). The corresponding point on the graph is
(a,f(a)) = (a,(a-2)^1/2). Hence the equation of the general tangent line is

y - f(a) = f'(a) ( x - a ),
i.e.,
\[ y - \sqrt{a-2} = \frac{1}{2\sqrt{a-2}} (x - a) \]
Now this equation goes through (5,2) if
\[ 2 - \sqrt{a-2} = \frac{1}{2\sqrt{a-2}} (5 - a) \]
Multiply both sides by \( 2\sqrt{a-2} \),
\[ 4\sqrt{a-2} - 2(a-2) = 5 - a \]
i.e.
\[ 4 \sqrt{a-2} = a + 1 \]

Now square both sides and solve for a.

Answer 24

JamesJ you are amazing. You just reminded me that I could have just used point-slope form.

Answer 25

@JamesJ - where did I go wrong?

Answer 26

was one of my assumptions incorrect?

Answer 27

@asnaseer, you were right. You just didn't use the form for y = f(x) in your last equation.

Answer 28

aaahhhh - ok - now I see - thanks for the clarification @JamesJ