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?
Thanks in advanced.
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
Any point on that graph looks like $\displaystyle (a,a^2-4a+25).$
So any tangent passing through the origin must have slope $\displaystyle \dfrac{a^2-4a+25}{a}$.
Using derivatives we know that the slope at any point is $\displaystyle 2a-4$.
Where are those equal?
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.