Hello there. To solve this question, we'll simply have to make x => x + 5 in the function.
Given the function:
[tex]f(x)=4x^2-3^{}[/tex]
We have to determine f(x + 5)
By making x => x + 5 in this function, we get:
[tex]f(x+5)=4\cdot(x+5)^2-3[/tex]
Now remember the binomial expansion of order 2:
[tex](a+b)^2=a^2+2ab+b^2[/tex]
Therefore we have:
[tex]f(x+5)=4\cdot(x^2+2\cdot x\cdot5+5^2)-3[/tex]
Multiply the terms inside parentheses and calculate the square.
[tex]f(x+5)=4\cdot(x^2_{}+10x+25)-3[/tex]
Apply the distributive property
[tex]f(x+5)=4x^2+4\cdot10x+4\cdot25-3[/tex]
Multiply and add the numbers
[tex]\begin{gathered} f(x+5)=4x^2+40x+100-3 \\ \boxed{f(x+5)=4x^2+40x+97} \end{gathered}[/tex]
This is the answer we're looking for.
A way of showing this is the correct answer is to make x = 1 and x = 6 in the former function:
[tex]\begin{gathered} f(1)=4\cdot1^2-3=4\cdot1-3=4-3=1 \\ f(6)=4\cdot6^2-3=4\cdot36-3=144-3=141 \end{gathered}[/tex]
Then making x = 1 in the expression we found after:
[tex]f(1+5)=f(6)=4\cdot1^2+40\cdot1+97=4+40+97=141[/tex]
As expected.
