Quadratic/Power Function Help

**jarny** · Nov 17th 2006, 03:44 PM

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!

**topsquark** · Nov 17th 2006, 04:20 PM

Originally Posted by jarny

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

**jarny** · Nov 17th 2006, 07:14 PM

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

**earboth** · Nov 17th 2006, 08:30 PM

Originally Posted by jarny

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

**Soroban** · Nov 18th 2006, 03:16 AM

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 $(8,2).$

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

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

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

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

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

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

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