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Find the given function values for a function $g$ where $g(x)=5x^2-3x+4\text{.}$

$g(3)=$

$g(0)=$

$g(-2)=$

Solution

We will substitute $x=3$ into $g(x)\text{:}$
$\begin{aligned} g(\substitute{3}) \amp= 5(\substitute{3})^2-3(\substitute{3})+4\\ \amp= 5(9)-9+4\\ \amp= 40 \end{aligned}$

We will substitute $x=0$ into $g(x)\text{:}$
$\begin{aligned} g(\substitute{0}) \amp= 5(\substitute{0})^2-3(\substitute{0})+4\\ \amp= 5(0)+0+4\\ \amp= 4 \end{aligned}$

We will substitute $x=-2$ into $g(x)\text{.}$ Especially when inputting negative numbers, be certain to put parentheses around the input values in the first step:
$\begin{aligned} g(\substitute{-2})\amp= 5(\substitute{-2})^2-3(\substitute{-2})+4\\ \amp= 5(4)+6+4\\ \amp= 30 \end{aligned}$

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