Problem:

$\quad \quad \quad \quad \tt{f(x) = { - x}^(2) + 1} \: \: when \: \: x = - 3$

Let's try!

$\quad \quad \quad \quad \tt{ ⟶f(x) = { - x}^(2) + 1}$

$\quad \quad \quad \quad \tt{⟶f(-3) = {- (-3)}^(2) + 1}$

$\quad \quad \quad \quad \tt{⟶ f( - 3) = { - (9) + 1}}$

$\quad \quad \quad \quad \tt{⟶ f( - 3) = { - 9 + 1}}$

$\quad \quad \quad \quad \tt{ ⟶f( - 3) = { -8}}$

$\quad \quad \quad \quad \boxed{\tt{ \color{green}f( - 3) = { - 8}}}$

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#LetsStudy

Question 3

$\longrightarrow{\green{A.\:-8}}$ ✔

Explanation:

$f(x )= - {x}^(2) + 1$

Plugging in the value "
$x\:=\:-3$ " in the above expression, we have

$f( - 3) = - ({ - 3})^(2) + 1 \\ \\ = - ( - 3 * - 3) + 1 \\ \\ = - (9) + 1 \\ \\ = - 9 + 1 \\ \\ = - 8$

$\bold{ \green{ \star{ \orange{Mystique35}}}}⋆$