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Math Help - Tangents

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by Adrienne View Post
    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-tē) 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}
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  3. #3
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    Quote Originally Posted by Adrienne View Post
    ...

    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)
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