I imagine is the point where the tangent cross the curve, doesn't it ?
Your method is good, but you can solve the equation if you notice that is solution, then put in factor..
I've got an equation: 4x^3 - 24x^2 + 45 x - 23
A tangent is drawn at point P. (1,2)
dy/dx = 12x^2 - 48x +45
At f´(1) = 12 -48 + 45 = 9
2 = 9(1) +c
y = 9x - 7
Show that the x coordinate of point q is 4.
9x-7 = 4x^3 - 24x^2 + 45 x - 23
0 = 4x^3 - 24x^2 + 36 x - 16
I found x = 4 using polyrootfinder on my gdc and was wondering if there is any other way of coming to that answer?
Of course !
You have to test some values such as 1, 2, 3, -1, -2, -3.
Indeed here : so we can put in factor :
which we develop :
And we identify each coefficient
After resolution we get :
Hence and you know the factorisation 2nd degree polynomials.
Is my english correct ?