# tangentes

• January 14th 2013, 07:19 AM
Boo
tangentes
if
f(x)=x^2-7x+6
and
g(x)= (x-1)(x^2+ax-2)

a)for which a afre graphs of the f and g functions in the (1,0) intersect under the angle of pi/4?

b) for which dots x tangent on graph of the function f in the point (x,f(x)) goes through the point (0,2)?

Many thanks!
• January 14th 2013, 08:02 AM
HallsofIvy
Re: tangentes
Quote:

Originally Posted by Boo
if
f(x)=x^2-7x+6
and
g(x)= (x-1)(x^2+ax-2)

a)for which a afre graphs of the f and g functions in the (1,0) intersect under the angle of pi/4?

f'(x)= 2x- 7 and $g'(x)= 1(x^2+ ax- 2)+ (x- 1)(2x+ a)= x^2+ ax- 2+ 2x^2+ (a- 2)x- a= 3x^2+ (2a- 2)x- (a+ 2)$
Those are, of course, the tangent of the angles they make with the x-axis: $tan(\theta_f)= 2x- 7$, $tan(\theta_g)= 3x^2+ (2a-2)- (a+2)$.

Now, use the trig identity $tan(\theta_f- \theta_g)= \frac{tan(\theta_f)- tan(\theta_g)}{1- tan(\theta_f)tan(\theta_g)}$

Quote:

b) for which dots x tangent on graph of the function f in the point (x,f(x)) goes through the point (0,2)?
As I said, f'(x)= 2x- 7 so a tangent line, at $(x_0, f(x_0))$, through (0, 2) must be of the form $y= (2x_0-7)x+ 2$. Set $(2x_0- 7)x_0+ 2= x_0^2- 7x_0+ 2$ and solve for $x_0$

[tex]Many thanks![/QUOTE]
• January 15th 2013, 03:56 AM
Boo
Re: tangentes
Hello, HallsofIvy!
then I get :
2x^2-7x+2=x^2-7x+2

and then:
x^2=0

the result should be T1(2,-4) and T2(-2, 24)

P.. HOw did U get x0 in latex?
Many thanks!!!