Question

Express f(x)=3x^2-2x+8 in the form of f(x)=a(x+b)^2+c

Answer 1

To express the function \( f(x) = 3x^2 - 2x + 8 \) in the form \( f(x) = a(x+b)^2 + c \), we will use the method of completing the square.

1. **Factor out the coefficient of \( x^2 \)** from the quadratic terms:
\[
f(x) = 3\left(x^2 - \frac{2}{3}x\right) + 8
\]

2. **Complete the square** for the expression inside the parentheses:
- Take the coefficient of \( x \) from \( x^2 - \frac{2}{3}x \), which is \( -\frac{2}{3} \).
- Half of this coefficient is \( -\frac{1}{3} \).
- Squaring this gives \( \left(-\frac{1}{3}\right)^2 = \frac{1}{9} \).

3. **Add and subtract this square** inside the parentheses:
\[
f(x) = 3\left(x^2 - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9}\right) + 8
\]
\[
f(x) = 3\left(\left(x - \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 8
\]

4. **Distribute \( 3 \)**:
\[
f(x) = 3\left(x - \frac{1}{3}\right)^2 - 3 \cdot \frac{1}{9} + 8
\]
\[
f(x) = 3\left(x - \frac{1}{3}\right)^2 - \frac{1}{3} + 8
\]
\[
f(x) = 3\left(x - \frac{1}{3}\right)^2 + \left(8 - \frac{1}{3}\right)
\]

5. **Combine the constants**:
\[
8 = \frac{24}{3}
\]
\[
8 - \frac{1}{3} = \frac{24}{3} - \frac{1}{3} = \frac{23}{3}
\]

Now we can express \( f(x) \) in the completed square form:
\[
f(x) = 3\left(x - \frac{1}{3}\right)^2 + \frac{23}{3}
\]

Thus, the function can be expressed as:
\[
f(x) = 3\left(x - \frac{1}{3}\right)^2 + \frac{23}{3}
\]
where \( a = 3 \), \( b = -\frac{1}{3} \), and \( c = \frac{23}{3} \).