Two tangent lines...

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vileoxidation

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Oct 23, 2005
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Okay, need help with this problem:

Given f(x)=x^2-4x+5, find equations for two lines that are tangent to the graph of g that also pass through (0,1).

Help please! Thanks!
 
Let some point on parabola where the tangent to the curve passes through (0,1) be (p,q).

The slope of the tangent from (p,q) to (0,1) will be given by the derivative at x=0, and also by the slope equation (y2-y1)/(x2-1).

Equate and solve for p.
 
I don't understand at all...How am I supposed to find both tangents? A\nd why would the slope be the derivative at x=0?
 
Are you having difficulty solving the equation you've formed? Or are you having trouble forming the equation?
 
sorry to jump in but I think you need to define 2 points on your eqn. as the the tangent points (a,b) and (c,d).

Then y =x^2 -4x +5

then dy/dx = 2x -4
this is the slope of your tangent lines when x= a and x= c

Then use the known point of (0,1) with the gradient to form up the 2 equations required.
 
Okay, okay, now I am terribly terribly confused. What I have so far is f'(x)=2x-4, this is the slope of the two lines I need to find, and they both have to pass through (0,1).

But from here I am confused.
 
here is some more direction...
Let the 2 points on your curve be (a,b) and (c,d)

and f'(x)=2x-4

then at point (a,b)
f'(a) = 2a - 4

and at point (c,d)
f'(c) = 2c - 4

Now the general form of a straight line is y= mx + b and you have 2 sets of points and a slope - (a,b) & (0,1) m=2a - 4
and (c,d) & (0,1) m= 2c - 4.

Use this data to create your equations.
 
Ugh okay I am still not understanding. Now I have b=(2a-4)a+1 and the same equation for the point (c,d). But how do I find the slopes? I am totally missing something...
 
getting very close now, but don't forget the original equation is also of use to you. Put the point(a,b) into y = x^2 -4x +5

now you have 2 equations and 2 unknowns... back to you
 
Oh okay I think I got it. So are the two equations I should end up with y=1 and y=-8x+1?

Thank you!!
 
well done but need to think a little more... i could be wrong

take the first point (a,b)

b = (2a-4)a +1
b = 2a^2 -4a +1

and the orinal eqn...
y =x^2 -4x +5

b =a^2 -4a +5

let these equal each other
2a^2 -4a +1 = a^2 -4a +5

a^2 = 4
a=2 (could be plus or minus)

the to find b...
b =a^2 -4a +5
= 4 -8 +5
= 1
(but using a=-2 , b = 4 + 8 +5 and b= 17 giving (-2, 17) )

then using only (a,b) is ( 2,1)
back to y= (2a-4)x + 1

y = 1 and y=-8x+1 (for the other)

but still concerned about the posible point where a= -2 this gives
y= -8x + 1

let me think for a moment
 
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