Question

if $f(x) = 2x^2 - x + 3$, find the following.
$f(3) = $
$f(-3) = $
$f(a) = $
$f(-a) = $
$f(a + 1) = $
$2f(a) = $
$f(2a) = $
$f(a^2) = $
$f(a)^2 = $
$f(a + h) = $

Explanation:

For \( f(3) \):

Step1: Substitute \( x = 3 \) into \( f(x) \)

\( f(3)=2(3)^{2}-3 + 3 \)

Step2: Simplify the expression

First, calculate \( (3)^{2}=9 \), then \( 2\times9 = 18 \). So \( 18-3 + 3=18 \).

For \( f(-3) \):

Step1: Substitute \( x=-3 \) into \( f(x) \)

\( f(-3)=2(-3)^{2}-(-3)+3 \)

Step2: Simplify the expression

Calculate \( (-3)^{2}=9 \), \( 2\times9 = 18 \), \( -(-3)=3 \). Then \( 18 + 3+3 = 24 \).

For \( f(a) \):

Step1: Substitute \( x = a \) into \( f(x) \)

\( f(a)=2a^{2}-a + 3 \)

For \( f(-a) \):

Step1: Substitute \( x=-a \) into \( f(x) \)

\( f(-a)=2(-a)^{2}-(-a)+3 \)

Step2: Simplify the expression

Since \( (-a)^{2}=a^{2} \) and \( -(-a)=a \), we get \( f(-a)=2a^{2}+a + 3 \).

For \( f(a + 1) \):

Step1: Substitute \( x=a + 1 \) into \( f(x) \)

\( f(a + 1)=2(a + 1)^{2}-(a + 1)+3 \)

Step2: Expand and simplify

Expand \( (a + 1)^{2}=a^{2}+2a + 1 \), then \( 2(a^{2}+2a + 1)=2a^{2}+4a + 2 \). Then \( 2a^{2}+4a + 2-(a + 1)+3=2a^{2}+4a + 2 - a-1 + 3=2a^{2}+3a + 4 \).

For \( 2f(a) \):

Step1: First find \( f(a) \) and then multiply by 2

We know \( f(a)=2a^{2}-a + 3 \), so \( 2f(a)=2(2a^{2}-a + 3) \)

Step2: Distribute the 2

\( 2\times2a^{2}-2\times a+2\times3 = 4a^{2}-2a + 6 \)

For \( f(2a) \):

Step1: Substitute \( x = 2a \) into \( f(x) \)

\( f(2a)=2(2a)^{2}-(2a)+3 \)

Step2: Simplify the expression

Calculate \( (2a)^{2}=4a^{2} \), \( 2\times4a^{2}=8a^{2} \). So \( f(2a)=8a^{2}-2a + 3 \).

For \( f(a^{2}) \):

Step1: Substitute \( x = a^{2} \) into \( f(x) \)

\( f(a^{2})=2(a^{2})^{2}-a^{2}+3 \)

Step2: Simplify the expression

Since \( (a^{2})^{2}=a^{4} \), we get \( f(a^{2})=2a^{4}-a^{2}+3 \).

For \( [f(a)]^{2} \):

Step1: First find \( f(a) \) and then square it

We know \( f(a)=2a^{2}-a + 3 \), so \( [f(a)]^{2}=(2a^{2}-a + 3)^{2} \)

Step2: Expand the square

Using the formula \( (A + B + C)^{2}=A^{2}+B^{2}+C^{2}+2AB + 2AC+2BC \) where \( A = 2a^{2} \), \( B=-a \), \( C = 3 \).
\( (2a^{2})^{2}+(-a)^{2}+3^{2}+2(2a^{2})(-a)+2(2a^{2})(3)+2(-a)(3) \)
\( =4a^{4}+a^{2}+9-4a^{3}+12a^{2}-6a \)
\( =4a^{4}-4a^{3}+13a^{2}-6a + 9 \)

For \( f(a + h) \):

Step1: Substitute \( x=a + h \) into \( f(x) \)

\( f(a + h)=2(a + h)^{2}-(a + h)+3 \)

Step2: Expand and simplify

Expand \( (a + h)^{2}=a^{2}+2ah+h^{2} \), then \( 2(a^{2}+2ah + h^{2})=2a^{2}+4ah + 2h^{2} \). Then \( 2a^{2}+4ah + 2h^{2}-(a + h)+3=2a^{2}+4ah + 2h^{2}-a - h+3 \)

Answer:

s:
\( f(3)=\boldsymbol{18} \)

\( f(-3)=\boldsymbol{24} \)

\( f(a)=\boldsymbol{2a^{2}-a + 3} \)

\( f(-a)=\boldsymbol{2a^{2}+a + 3} \)

\( f(a + 1)=\boldsymbol{2a^{2}+3a + 4} \)

\( 2f(a)=\boldsymbol{4a^{2}-2a + 6} \)

\( f(2a)=\boldsymbol{8a^{2}-2a + 3} \)

\( f(a^{2})=\boldsymbol{2a^{4}-a^{2}+3} \)

\( [f(a)]^{2}=\boldsymbol{4a^{4}-4a^{3}+13a^{2}-6a + 9} \)

\( f(a + h)=\boldsymbol{2a^{2}+4ah + 2h^{2}-a - h + 3} \)