Chapter 14, Problem 1RE

To determine

To calculate: The value of function $f\left(x,y\right)=6y^2-5xy+2x$ at points $\left(-3,1\right)$ and $\left(5,-2\right).$

Answer to Problem 1RE

Solution:

The value of function $f\left(x,y\right)=6y^2-5xy+2x$ at points $\left(-3,1\right)$ and $\left(5,-2\right)$ is $\underset{\_}{15}$ and $\underset{\_}{84}$, respectively.

Explanation of Solution

Given:

The function $f\left(x,y\right)=6y^2-5xy+2x$ and points $\left(-3,1\right)$ and $\left(5,-2\right).$

Calculation:

In the function:

$f\left(x,y\right)=6y^2-5xy+2x$

To find the value of function at point $\left(-3,1\right)$, replace $x$ with $-3$ and $y$ with $1$,

$\begin{array}{ccc}\hfill f\left(-3,1\right)& =& 6{\left(1\right)}^2-5\left(-3\right)\left(1\right)+2\left(-3\right)\hfill \\ \hfill & =& 6+15-6\hfill \\ \hfill & =& 15\hfill \end{array}$

To find the value of function at point $\left(5,-2\right)$, replace $x$ with $5$ and $y$ with $-2$,

$\begin{array}{ccc}\hfill f\left(5,-2\right)& =& 6{\left(-2\right)}^2-5\left(5\right)\left(-2\right)+2\left(5\right)\hfill \\ \hfill & =& 24+50+10\hfill \\ \hfill & =& 84\hfill \end{array}$