If a function `f (x)` is given as `f (x) = x^(2) -3x +2` for all `x in R,` then `f (a +h)=`

Options

A. `a ^(2) + (2a + 3) h -3a +2+ h ^(2)`
B. `a ^(2) + (2a + 3 ) h + 3a + 2 + h^(2)`
C. `a ^(2) + (2a -3) h + 3a + 2+ h^(2)`
D. `a ^(2) + (2a + 3) h + 3a + 2+h ^(2)`

Answer

To find \( f(a + h) \) for the function \( f(x) = x^2 - 3x + 2 \), we will substitute \( a + h \) into the function.

### Step-by-step Solution:

1. **Substitute \( a + h \) into the function:**
\[ f(a + h) = (a + h)^2 - 3(a + h) + 2 \]

2. **Expand \( (a + h)^2 \):**
\[ (a + h)^2 = a^2 + 2ah + h^2 \]
So, we have:
\[ f(a + h) = a^2 + 2ah + h^2 - 3(a + h) + 2 \]

3. **Distribute the \(-3\) in the expression:**
\[ -3(a + h) = -3a - 3h \]
Now substituting this back, we get:
\[ f(a + h) = a^2 + 2ah + h^2 - 3a - 3h + 2 \]

4. **Combine like terms:**
\[ f(a + h) = a^2 - 3a + 2 + 2ah + h^2 - 3h \]
Here, \( a^2 - 3a + 2 \) is a constant term, and the remaining terms involve \( h \).

5. **Final expression:**
\[ f(a + h) = (a^2 - 3a + 2) + (2a - 3)h + h^2 \]

### Final Result:
\[ f(a + h) = (a^2 - 3a + 2) + (2a - 3)h + h^2 \]