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  1. #1
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    calc 1

    find 2nd degree polynomial f(x) = ax^2+bx+c such that its graph has a tangent line with slope=6 @ point(-2, 7) and x-int @ (-1, 0).
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  2. #2
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    Quote Originally Posted by weezie23 View Post
    find 2nd degree polynomial f(x) = ax^2+bx+c such that its graph has a tangent line with slope=6 @ point(-2, 7) and x-int @ (-1, 0).
    We can use the points we know to find a system of equations

    $\displaystyle f(x) = ax^2+bx+c \to f'(x)=2ax+b$

    Since we know the derivative represents slope we know when x=-2 that $\displaystyle f'(-2)=6$ so we get

    $\displaystyle E_1: \\\ 6=-4a+b$

    Now using the point (-2,7) in f(x) we get

    $\displaystyle E_2: \\\ 7=4a-2b+c$

    and the point (-1,0) we get

    $\displaystyle E_3: \\\ 0=a-b+c$

    We now have a system of equations in a,b, and c

    Multiplying 3 by -1 and adding it to 2 gives

    $\displaystyle 7=3a-b$ Now if add equation one to this one we get

    $\displaystyle 13=-a \iff a=-13$

    Plugging this in above gives

    $\displaystyle 7=3(-13)-b \iff b=-46$

    Now putting a and b into equation 3 gives

    $\displaystyle 0=-13-(-46)+c \iff c=-33$

    So we get
    $\displaystyle f(x)=-13x^2-46x-33$

    I hope this helps.
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  3. #3
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    Hello, weezie23!

    Find a 2nd degree polynomial: $\displaystyle f(x) \:= \:ax^2+bx+c$ such that its graph
    has a tangent line with slope = 6 at point(-2, 7) and x-int (-1, 0)

    We have: .$\displaystyle f(x) \:=\:ax^2+bx + c$ .and we must find $\displaystyle a,b,c.$


    The point (-2,7) is on the graph.
    . . $\displaystyle 7 \:=\:a(-2)^2 + b(-2) + c\quad\Rightarrow\quad 4a - 2b + c \:=\:7\;\;{\color{blue}[1]}$

    The point (-1,0) is on the graph.
    . . $\displaystyle 0 \:=\:a(-1)^2 + b(-1) + c\quad\Rightarrow\quad a - b + c \:=\:-1\;\;{\color{blue}[2]}$

    When $\displaystyle x = -2$, the slope is 6.
    We have: .$\displaystyle f'(x) \:=\:2ax + b$
    . . $\displaystyle 6 \:=\:2a(-2) + b \quad\Rightarrow\quad -4a + b \:=\:6\;\;{\color{blue}[3]}$


    Solve the system of equations: .$\displaystyle \begin{array}{cccc}4a - 2b + c &=& 7 & {\color{blue}[1]} \\ a - b + c &=& 0 & {\color{blue}[2]} \\ -4a + b &=& 6 & {\color{blue}[3]} \end{array}$

    $\displaystyle \begin{array}{cccc}\text{Subtract {\color{blue}[2]} from {\color{blue}[1]}} & 3x - b &=&7 \\ \text{Add {\color{blue}[3]}} & \text{-}4a + b &=& 6 \end{array}$

    And we have: .$\displaystyle -a \:=\:13\quad\Rightarrow\quad \boxed{a \:=\:-13}$

    Substitute into [3]: .$\displaystyle -4(-13) + b \:=\:6 \quad\Rightarrow\quad\boxed{ b \:=\:-46}$

    Substitute into [2]: .$\displaystyle -13 + 46 + c \:=\:0 \quad\Rightarrow\quad\boxed{ c \:=\:-33}$


    Therefore: .$\displaystyle {\color{red}f(x) \;=\;-13x^2 - 46x - 33}$

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