# Math Help - Equations of all tangents..

1. ## Equations of all tangents..

Find the equations of all the tangents to the graph of (fx) = x^2 - 4x + 25 that passes through the origin.

Would I just use x=0?

2. No I don't think that is going to work.

You need the tangent going through the origin which is y-0= m (x-0) (or for simplicity's sake y=mx) where m = f'(a) where a is the point in x where the tangent occurs.

Find a the point where the line(s) and curve intersect

3. Any point on that graph looks like $(a,a^2-4a+25).$
So any tangent passing through the origin must have slope $\dfrac{a^2-4a+25}{a}$.

Using derivatives we know that the slope at any point is $2a-4$.

Where are those equal?

Find the equations of all the tangents to the graph of (fx) = x^2 - 4x + 25 that passes through the origin.
looks like there are two ...

5. thanks for the help everyone..

So I did this:

2x - 4

f'(0) = -4

y=(0)^2 - 4(0) + 25
y = 25

y=mx+b
25 = 14(0) +b
b= 25

And I'm not sure where I'm going..

thanks for the help everyone..

So I did this:

2x - 4

f'(0) = -4

y=(0)^2 - 4(0) + 25
y = 25

y=mx+b
25 = 14(0) +b
b= 25

And I'm not sure where I'm going..
Neither am I, why did you choose these calcs?

7. I thought I was to take the deriviate and find the slope of the tangent line first? (which would be -4)
and then use that to find the equation of the tangent which would be using b=25?

8. Your two tangents occur if $x=5~\&~x=-5.$
Now you reply with reasons why!

9. I don't know.. I am very new and confused to calculus

I don't know.. I am very new and confused to calculus

I understand it can be confusing

Originally Posted by Plato
Any point on that graph looks like $(a,a^2-4a+25).$
So any tangent passing through the origin must have slope $\dfrac{a^2-4a+25}{a}$.

Using derivatives we know that the slope at any point is $2a-4$.

Where are those equal?
Now solve $\dfrac{a^2-4a+25}{a}=2a-4$

Can you do that?

11. note that a curve and a line tangent to the curve have the following two properties ...

1) their slopes are equal

2) the curve and tangent line are equal at the point of tangency

using property (1), m = 2x-4

a line through the origin has the form y = mx , so the line may be written y = (2x-4)x

now, use property (2) to solve for the values of x where the curve = the tangent line.

12. I am so sorry, I feel like I'm wasting everyone's time but I am really lost.
I don't know how to respond, but I really appreciate your time and sincere ways of trying to help.. Im trying to do more research on the side to figure this out