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Expert-verified Found in: Page 156 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # Find the value of $k\left(0\right),k\left(2\right),k\left(-2\right),k\left(\sqrt{2}\right),k\left(a+2\right),k\left(-x\right),k\left({x}^{2}\right)$ for the function $k\left(x\right)=-{x}^{2}-2x+3$.

The value of function are

$\begin{array}{l}k\left(0\right)=3,k\left(2\right)=-5,k\left(-2\right)=3,k\left(\sqrt{2}\right)=1-2\sqrt{2},k\left(a+2\right)=-{a}^{2}-6a-5,k\left(-x\right)=-{x}^{2}+2x+3,\\ k\left({x}^{2}\right)=-{x}^{4}-2{x}^{2}+3\end{array}$.

See the step by step solution

## Step 1. Evaluate k0,k2,k−2,k2 for the given function.

From the given function,

$k\left(x\right)=-{x}^{2}-2x+3....\left(1\right)$

Evaluate the value of function at $x=0,2,-2$ gives,

role="math" localid="1644504817782" $k\left(0\right)=-{\left(0\right)}^{2}-2\left(0\right)+3....\left(\text{Substitute}x=0\text{in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=3\phantom{\rule{0ex}{0ex}}k\left(2\right)=-{\left(2\right)}^{2}-2\left(2\right)+3....\left(\text{Substitute}x=2\text{\hspace{0.17em}in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=-5\phantom{\rule{0ex}{0ex}}k\left(-2\right)=-{\left(-2\right)}^{2}-2\left(-2\right)+3....\left(\text{Substitute}x=-2\text{in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=3\phantom{\rule{0ex}{0ex}}k\left(\sqrt{2}\right)=-{\left(\sqrt{2}\right)}^{2}-2\left(\sqrt{2}\right)+3....\left(\text{Substitute}x=\sqrt{2}\text{in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=1-2\sqrt{2}$

## Step 2. Evaluate ka+2,k−x,kx2 for the given function.

Similarly, evaluate the value of function (1) at $x=a+2,-x,{x}^{2}$ gives,

role="math" localid="1644505036706" $k\left(a+2\right)=-{\left(a+2\right)}^{2}-2\left(a+2\right)+3....\left(\text{Substitute}x=a+2\text{in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=-{a}^{2}-4a-4-2a-4+3\phantom{\rule{0ex}{0ex}}=-{a}^{2}-6a-5\phantom{\rule{0ex}{0ex}}k\left(-x\right)=-{\left(-x\right)}^{2}-2\left(-x\right)+3....\left(\text{Substitute}x=-x\text{\hspace{0.17em}in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=-{x}^{2}+2x+3\phantom{\rule{0ex}{0ex}}k\left({x}^{2}\right)=-{\left({x}^{2}\right)}^{2}-2\left({x}^{2}\right)+3....\left(\text{Substitute}x={x}^{2}\text{in}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=-{x}^{4}-2{x}^{2}+3$

## Step 3. Description of steps.

From the obtained values,

$\begin{array}{l}k\left(0\right)=3,k\left(2\right)=-5,k\left(-2\right)=3,k\left(\sqrt{2}\right)=1-2\sqrt{2},k\left(a+2\right)=-{a}^{2}-6a-5,k\left(-x\right)=-{x}^{2}+2x+3,\\ k\left({x}^{2}\right)=-{x}^{4}-2{x}^{2}+3\end{array}$. ### Want to see more solutions like these? 