Answer
(a)
$\begin{cases} 100a + 10b + c=25 \\ 225a + 15b + c=33.75 \\1600a + 40b + c=40\end{cases}$
(b)
$a=-0.05$
$b=3$
$c=0$
$y=-0.05x^2+3x$
Work Step by Step
(a) Use the figure given in the Exercise, the surveyed points are $(10,25),(15,33.75),(40,40)$. Use these points with the model equation $y=ax^2+bx+c$, we have
$\begin{cases} 100a + 10b + c=25 \\ 225a + 15b + c=33.75 \\1600a + 40b + c=40\end{cases}$
(b) To solve the above equations with Cramer's Rule, set up the following determinants:
$\begin{array}( \\|D|= \\ \\ \end{array}\begin{vmatrix} 100&10&1\\225&15&1\\1600&40&1 \end{vmatrix},
\begin{array}( \\|D_a|= \\ \\ \end{array}\begin{vmatrix} 25&10&1\\33.75&15&1\\40&40&1 \end{vmatrix},\begin{array}( \\|D_b|= \\ \\ \end{array}\begin{vmatrix} 100&25&1\\225&33.75&1\\1600&40&1 \end{vmatrix},\begin{array}( \\|D_c|= \\ \\ \end{array}\begin{vmatrix} 100&10&25\\225&15&33.75\\1600&40&40 \end{vmatrix}$
Do the following operations for the first three determinants: $R_2-R_1\to R_2, R_3-R_1\to R_3$, and
do $2R_2-3R_1\to R_2, R_3-4R_1\to R_3$ for the last one.
$\begin{array}( \\|D|= \\ \\ \end{array}\begin{vmatrix} 100&10&1\\125&5&0\\1500&30&0 \end{vmatrix},
\begin{array}( \\|D_a|= \\ \\ \end{array}\begin{vmatrix} 25&10&1\\8.75&5&0\\15&30&0 \end{vmatrix},\begin{array}( \\|D_b|= \\ \\ \end{array}\begin{vmatrix} 100&25&1\\125&8.75&0\\1500&15&0 \end{vmatrix},\begin{array}( \\|D_c|= \\ \\ \end{array}\begin{vmatrix} 100&10&25\\150&0&-7.5\\1200&0&-60 \end{vmatrix}$
Evaluate the above determinants using the columns with two zeros:
$|D|=125\times30-5\times1500=-3750$
$|D_a|=8.75\times30-5\times15=187.5$
$|D_b|=125\times15-8.75\times1500=-11250$
$|D_c|=-10(150(-60)+7.5\times1200)=0$
We get the final results as:
$a=\frac{|D_a|}{|D|}=-0.05$
$b=\frac{|D_b|}{|D|}=3$
$c=\frac{|D_c|}{|D|}=0$
and the model function is $y=-0.05x^2+3x$
