Question

Can I please get answer for a and b. I just need the answers. Thanks

Answer 1

When we have a function and want to evalueate it for some "x", we just substitute every "x" with the value in parenthesis. So, to evaluate f(a + 2), we have x = a + 2, so we need to substitute every "x" in the function with a + 2:

`f(a+2)=(a+2)^2-2(a+2)+1`

Now, to simplify, we first need to use the distributive property on the second parenthesis and evaluate the binomial in the first:

`f(a+2)=a^2+4a+4-2a-4+1`

Now, we group the like-terms and add them:

`\begin{gathered} f(a+2)=a^2+4a-2a+4+4+1 \\ f(a+2)=a^2+2a+1 \end{gathered}`

For yhe other one, Let's do each term and then make the substraction:

`\begin{gathered} f(a+h)=(a+h)^2-2(a+h)+1 \\ f(a+h)=a^2+2ah+h^2-2a-2h+1 \\ f(a+h)=a^2+h^2+2ah-2a-2h+1 \end{gathered}`
`f(a)=a^2-2a+1`

So:

`\begin{gathered} f(a+h)-f(a)=a^2+h^2+2ah-2a-2h+1-(a^2-2a+1) \\ f(a+h)-f(a)=a^2-a^2+h^2+2ah-2a+2a-2h+1-1 \\ f(a+h)-f(a)=h^2+2ah-2h \end{gathered}`