# Tangents

• Oct 29th 2007, 09:46 PM
Tangents
Hey guys

I have an Calc test tomorrow and I had some trouble when reviewing with these few questions. Please help me out with them? :]

1) Find the equations of all the tangent lines to the graph of f(x)=4x-x^2 that pass through the point (2,5)

2) The line y=x-6 is parallel to the tangent drawn to the curve y= (kx+8)/(k+x) at x=-2. What is the value of k?

3) If p(x)=(x-1)(x+k) and if the line tangent to the graph of p at the point (4, p(4)) is parallel to the line 5x-y+6=0, then find the value of k

4) Let f be the function f(x)=x^3+3x^2-x+2. The tangent to the graph of f at the point P(2, f(2)) intersects the graph of f again at point q. Find the coordinates of Q

Thanks a bundle!
• Oct 29th 2007, 10:43 PM
earboth
Quote:

Originally Posted by Adrienne
Hey guys

...
1) Find the equations of all the tangent lines to the graph of f(x)=4x-x^2 that pass through the point (2,5)

2) The line y=x-6 is parallel to the tangent drawn to the curve y= (kx+8)/(k+x) at x=-2. What is the value of k?

...

Hello,

to 1): The equation of a straight line is: y = mx + b. The slope m must be the same as the 1rt derivative of f if this line is a tangent to the graph of f.
Let T(t, 4t-t²) be the tangent point then

$m=f'(t)=4-2t$ . Use point-slope-formula to calculate the equation of the tangent at T:

$\frac{y-(4t-t^2)}{x-t}=4-2t$ After a few steps of rearranging you'll get:

$y = (4-2t)x+t^2$

The given point is placed on this tangent too that means it's coordinate must satisfy this equation too. Plug in the coordinates:

$5=(4-2t) \cdot 2 + t^2$ and solve for t. I've got t= 1 or t = 3

Plug in these values of t into the equation of the tangent:

y = 2x + 1 or y = -2x + 9

to 2): Given is $f_k(x) = \frac{kx+8}{k+x}$ . If a tangent to the graph of $f_k$ is parallel to y = x - 6 then the 1rt derivative equals 1:

$f_k'(x) = \frac{(k+x) \cdot k - (kx+8)}{(k+x)^2}$ . With x = -2 you get the equation:

$\frac{(k-2) \cdot k - (-2k+8)}{(k-2)^2}=1~\iff~ \frac{k^2-8}{(k-2)^2}=1$ After a few steps of rearranging you'll get k = 3.
Plug in this value into the equation of $f_k$ and you'll get

$f_{3}(x) = \frac{3x+8}{3+x}$
• Oct 29th 2007, 11:05 PM
earboth
Quote:

Originally Posted by Adrienne
...

3) If p(x)=(x-1)(x+k) and if the line tangent to the graph of p at the point (4, p(4)) is parallel to the line 5x-y+6=0, then find the value of k

4) Let f be the function f(x)=x^3+3x^2-x+2. The tangent to the graph of f at the point P(2, f(2)) intersects the graph of f again at point q. Find the coordinates of Q

...

Hi,

to 3):
1. Expand the equation of p: $p_k(x)=x^2-x+kx-k~\implies~ p_k'(x)=2x-1+k$
2. Plug in the x-coordinate of the tangent point: $p_k'(4)=8-1+k$
3. This value equals 5 which is the slope of the given line: $7+k=5~\implies~k=-2$
4. Plug in this value into the equation of $p_k$: $p_{-2}(x) = (x-1)(x-2)$

to 4)
1. f(2)=20
2. $f'(x) = 3x^2+6x-1~\implies~f'(2)=23$
3. Use point-slope-formula of a line to calculate the equation of the tangent:

$\frac{y-20}{x-2}=23~\implies~y=23x-26$

4. Calculate the intersection:

$x^3+3x^2-x+2=23x-26~\iff~x^3+3x^2-24x+28=0$

5. x = 2 must be a solution of this equation. Use long division to reduce the equation:

$\frac{x^3+3x^2-24x+28}{x-2}= x^2+5x-14$

6. Now solve $x^2+5x-14=0~\iff~(x+7)(x-2)=0$

7. As you see the intersection you are looking for is at Q(-7, -187)