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Math Help - Quadratic/Power Function Help

  1. #1
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    Quadratic/Power Function Help

    How do you find the equation for the power function y=kx^a with points (3,15) and (8,2). Also how do you find the quadratic equation in standard form containing points (3,26.6), (8,36.6), and (14, 13.4)? Help would be appreciated thank you!
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  2. #2
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    Quote Originally Posted by jarny View Post
    How do you find the equation for the power function y=kx^a with points (3,15) and (8,2). Also how do you find the quadratic equation in standard form containing points (3,26.6), (8,36.6), and (14, 13.4)? Help would be appreciated thank you!
    For the first, take the log of both sides:
    log(y) = a \cdot log(x) + log(k)

    So:
    log(15) = a \cdot log(3) + log(k)
    log(2) = a \cdot log(8) + log(k)
    is a system of two equations in two unknowns. So solve for a and log(k).

    For the second, the standard form is:
    y = ax^2 + bx + c

    You have three points, so that will give you three equations in three unknowns: a, b, and c.

    -Dan
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  3. #3
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    Still having a little trouble.....

    Im still having a little trouble understanding the power function. The quadratic i get but could someone post the solution/way to do it so i can check my work? thanks
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  4. #4
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    Quote Originally Posted by jarny View Post
    Im still having a little trouble understanding the power function. The quadratic i get but could someone post the solution/way to do it so i can check my work? thanks
    Hello, Jarny,

    I take over topsquark's system of equation:
    \log(15) = a \cdot \log(3) + \log(k)
    \log(2) = a \cdot \log(8) + \log(k)
    is a system of two equations in two unknowns. So solve for a and log(k).
    From the first equation you get:

    \log(15) - a \cdot \log(3) = \log(k)

    Plug in \log(k) into the 2nd equation:

    \log(2) = a \cdot \log(8) + \log(15) - a \cdot \log(3)

    Rearrange with the variable a on one side of the equation:

    \log(2) -  \log(15)= a \cdot \log(8)  - a \cdot \log(3)

    \log(2) -  \log(15)= a \cdot \left(\log(8)  -  \log(3)\right)

    \frac{\log(2) -  \log(15)}{\log(8)  -  \log(3)}= a

    \frac{\log \left( \frac{2}{15} \right)}{\log \left(\frac{8}{3} \right)}= a

    You can rewrite the last equation to:

    a=\log_{\frac{8}{3}}{\left( \frac{2}{15} \right)}

    EB
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  5. #5
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    Hello, jarny!

    I assume you could use a walk-through ... with baby-steps.
    . . Okay, here we go . . .


    Find the equation for the power function y=kx^a with points (3,15) and (8,2).

    \begin{array}{cc}\text{From}\\ \text{From}\end{array}<br />
\begin{array}{cc}(3,15), \\ (8,2),\end{array}<br />
\begin{array}{cc}\text{we have: } \\ \text{we have: }\end{array}<br />
\begin{array}{cc}k\cdot3^a\:=\:15 \\ k\cdot8^a\:=\:2\end{array}<br />
\begin{array}{cc}(1)\\(2)\end{array}

    Divide (1) by (2): . \frac{k\cdot3^a}{k\cdot8^a}\:=\:\frac{15}{2}\quad\  Rightarrow\quad\left(\frac{3}{8}\right)^a\:=\:\fra  c{15}{2}\quad\Rightarrow\quad (0.375)^a\:=\:7.5

    Take logs: . \ln(0.375)^a\:=\:\ln(7.5)\quad\Rightarrow\quad a\cdot\ln(0.375)\:=\:\ln(7.5)

    Hence: . a\:=\:\frac{\ln(7.5)}{\ln(0.375)}\quad\Rightarrow\  quad\boxed{ a\:\approx\:-2.054}

    Substitute into (1): . k\cdot3^{-2.054}\:=\:15\quad\Rightarrow\quad k \:=\;15\cdot3^{2.054}\quad\Rightarrow\quad\boxed{ k \:\approx\:143.25}


    Therefore, the power function is: . \boxed{y \;= \;143.25x^{-2.054}}

    . . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Check

    \begin{array}{cc}x=3:\\ x=8:\end{array}<br />
\begin{array}{cc}143.25(3^{-2.054})\:=\\143.25(8^{-2.054}) \:=\end{array}<br />
\begin{array}{cc}14.9998727 \\ 2.00054237\end{array} . close enough!



    Find the quadratic equation containing points (3,\,26.6),\;(8,\,36.6),\;(14,\,13.4)

    The quadratic function is of the form: . y \;= \;ax^2 + bx + c

    \begin{array}{cc}\text{From }(3,\,26.6) \\ \text{From }(8,\,36.6) \\ \text{From }(14,\,13.4)\end{array}<br />
\begin{array}{ccc}\text{we have:} \\ \text{we have:} \\ \text{we have:}\end{array}<br />
\begin{array}{ccc}9a + 3b + c\\64a + 8b + c\\196a + 14b + c\end{array}<br />
\begin{array}{ccc} =\:26.6\\ =\:36.6 \\ =\:13.4 \end{array}<br />
\begin{array}{ccc}(1)\\(2)\\(3)\end{array}

    \begin{array}{cc}\text{Subtract (1) from (2):} \\ \text{Subtract (2) from (3):}\end{array}<br />
\begin{array}{cc}55a + 5b \\132a + 6b\end{array}<br />
\begin{array}{cc} = \\ = \end{array}\!\!<br />
\begin{array}{cc}10\\\text{-}23.2\end{array} . <br />
\begin{array}{cc}\Rightarrow \\ \Rightarrow\end{array}<br />
\begin{array}{cc}11a + b\\66a + 3b\end{array}<br />
\begin{array}{cc} = \\ = \end{array}<br />
\begin{array}{cc}2 \\ \text{-}11.6\end{array}<br />
\begin{array}{cc}(4) \\ (5)\end{array}


    Multiply (4) by \text{-}3:\;\;\text{-}33a - 3b\:=\:\text{-}6
    . . . . . . . Add (5): . 66a + 3b \:= \:\text{-}11.6

    And we have: . 33a \:= \:\text{-}17.6\quad\Rightarrow\quad\boxed{ a \,= \,\text{-}\frac{8}{15}}

    Substitute into (4): . 11\left(\text{-}\frac{8}{15}\right) + b \:=\:2\quad\Rightarrow\quad\boxed{ b = \frac{118}{15}}

    Substitute into (1): . 9\left(\text{-}\frac{8}{15}\right) + 3\left(\frac{118}{15}\right) + c \:= \:26.6\quad\Rightarrow\quad\boxed{ c = \frac{39}{5}}


    Therefore: . \boxed{y \; = \;\text{-}\frac{8}{15}x^2 + \frac{118}{15}x + \frac{39}{5}} . or . \boxed{y \;= \;\text{-}\frac{1}{15}\left(8x^2 - 118x - 117\right)}

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