Quadratic Function Worksheets - Free Printable

To solve the problem, we need to convert each given quadratic function into its general form, which is:

\[
f(x) = ax^2 + bx + c
\]

where \(a\), \(b\), and \(c\) are constants, and \(a \neq 0\).

Let's go through each problem step by step.

---

Problem 1:

\[
f(x) = 6x^2 + 5
\]

This is already in general form:
\[
a = 6, \quad b = 0, \quad c = 5
\]

General Form:
\[
f(x) = 6x^2 + 0x + 5
\]

---

Problem 2:

\[
f(x) = -3\left(x + \frac{2}{3}\right)\left(x + 1\right)
\]

First, expand the expression:
\[
f(x) = -3\left(x + \frac{2}{3}\right)\left(x + 1\right)
\]

Use the distributive property (FOIL method):
\[
\left(x + \frac{2}{3}\right)\left(x + 1\right) = x^2 + x + \frac{2}{3}x + \frac{2}{3}
\]
\[
= x^2 + \left(1 + \frac{2}{3}\right)x + \frac{2}{3}
\]
\[
= x^2 + \frac{5}{3}x + \frac{2}{3}
\]

Now multiply by \(-3\):
\[
f(x) = -3\left(x^2 + \frac{5}{3}x + \frac{2}{3}\right)
\]
\[
= -3x^2 - 5x - 2
\]

General Form:
\[
f(x) = -3x^2 - 5x - 2
\]

---

Problem 3:

\[
f(x) = -(x - 6)^2
\]

Expand \((x - 6)^2\):
\[
(x - 6)^2 = x^2 - 12x + 36
\]

Now multiply by \(-1\):
\[
f(x) = -(x^2 - 12x + 36)
\]
\[
= -x^2 + 12x - 36
\]

General Form:
\[
f(x) = -x^2 + 12x - 36
\]

---

Problem 4:

\[
f(x) = 6x^2 + 7x - 120
\]

This is already in general form:
\[
a = 6, \quad b = 7, \quad c = -120
\]

General Form:
\[
f(x) = 6x^2 + 7x - 120
\]

---

Problem 5:

\[
f(x) = 2\left(x - \frac{3}{2}\right)^2
\]

Expand \(\left(x - \frac{3}{2}\right)^2\):
\[
\left(x - \frac{3}{2}\right)^2 = x^2 - 3x + \frac{9}{4}
\]

Now multiply by 2:
\[
f(x) = 2\left(x^2 - 3x + \frac{9}{4}\right)
\]
\[
= 2x^2 - 6x + \frac{9}{2}
\]

General Form:
\[
f(x) = 2x^2 - 6x + \frac{9}{2}
\]

---

Problem 6:

\[
f(x) = -x^2 + 9x + 2
\]

This is already in general form:
\[
a = -1, \quad b = 9, \quad c = 2
\]

General Form:
\[
f(x) = -x^2 + 9x + 2
\]

---

Problem 7:

\[
f(x) = -(x - 5)^2 - 1
\]

Expand \((x - 5)^2\):
\[
(x - 5)^2 = x^2 - 10x + 25
\]

Now multiply by \(-1\) and subtract 1:
\[
f(x) = -(x^2 - 10x + 25) - 1
\]
\[
= -x^2 + 10x - 25 - 1
\]
\[
= -x^2 + 10x - 26
\]

General Form:
\[
f(x) = -x^2 + 10x - 26
\]

---

Problem 8:

\[
f(x) = \frac{1}{2}(x - 3)(x + 9)
\]

Expand \((x - 3)(x + 9)\):
\[
(x - 3)(x + 9) = x^2 + 9x - 3x - 27
\]
\[
= x^2 + 6x - 27
\]

Now multiply by \(\frac{1}{2}\):
\[
f(x) = \frac{1}{2}(x^2 + 6x - 27)
\]
\[
= \frac{1}{2}x^2 + 3x - \frac{27}{2}
\]

General Form:
\[
f(x) = \frac{1}{2}x^2 + 3x - \frac{27}{2}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad f(x) = 6x^2 + 0x + 5 \\
2. & \quad f(x) = -3x^2 - 5x - 2 \\
3. & \quad f(x) = -x^2 + 12x - 36 \\
4. & \quad f(x) = 6x^2 + 7x - 120 \\
5. & \quad f(x) = 2x^2 - 6x + \frac{9}{2} \\
6. & \quad f(x) = -x^2 + 9x + 2 \\
7. & \quad f(x) = -x^2 + 10x - 26 \\
8. & \quad f(x) = \frac{1}{2}x^2 + 3x - \frac{27}{2}
\end{aligned}
}
\]