Quadratic Function Worksheets - Free Printable

To solve the problem, we need to complete each function table by substituting the given values of \( x \) into the provided functions and calculating the corresponding values of \( y \). Let's go through each function step by step.

---

1. \( f(x) = -3x^2 + 6x - 19 \)

The table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-2 & \\
-1 & \\
0 & \\
1 & \\
2 & \\
\end{array}
\]

#### Substituting \( x = -2 \):
\[
f(-2) = -3(-2)^2 + 6(-2) - 19
\]
\[
= -3(4) + 6(-2) - 19
\]
\[
= -12 - 12 - 19
\]
\[
= -43
\]

#### Substituting \( x = -1 \):
\[
f(-1) = -3(-1)^2 + 6(-1) - 19
\]
\[
= -3(1) + 6(-1) - 19
\]
\[
= -3 - 6 - 19
\]
\[
= -28
\]

#### Substituting \( x = 0 \):
\[
f(0) = -3(0)^2 + 6(0) - 19
\]
\[
= -19
\]

#### Substituting \( x = 1 \):
\[
f(1) = -3(1)^2 + 6(1) - 19
\]
\[
= -3(1) + 6(1) - 19
\]
\[
= -3 + 6 - 19
\]
\[
= -16
\]

#### Substituting \( x = 2 \):
\[
f(2) = -3(2)^2 + 6(2) - 19
\]
\[
= -3(4) + 6(2) - 19
\]
\[
= -12 + 12 - 19
\]
\[
= -19
\]

The completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-2 & -43 \\
-1 & -28 \\
0 & -19 \\
1 & -16 \\
2 & -19 \\
\end{array}
\]

---

2. \( g(x) = -4x^2 + 32x - 63 \)

The table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & \\
0 & \\
1 & \\
2 & \\
3 & \\
\end{array}
\]

#### Substituting \( x = -1 \):
\[
g(-1) = -4(-1)^2 + 32(-1) - 63
\]
\[
= -4(1) + 32(-1) - 63
\]
\[
= -4 - 32 - 63
\]
\[
= -99
\]

#### Substituting \( x = 0 \):
\[
g(0) = -4(0)^2 + 32(0) - 63
\]
\[
= -63
\]

#### Substituting \( x = 1 \):
\[
g(1) = -4(1)^2 + 32(1) - 63
\]
\[
= -4(1) + 32(1) - 63
\]
\[
= -4 + 32 - 63
\]
\[
= -35
\]

#### Substituting \( x = 2 \):
\[
g(2) = -4(2)^2 + 32(2) - 63
\]
\[
= -4(4) + 32(2) - 63
\]
\[
= -16 + 64 - 63
\]
\[
= -15
\]

#### Substituting \( x = 3 \):
\[
g(3) = -4(3)^2 + 32(3) - 63
\]
\[
= -4(9) + 32(3) - 63
\]
\[
= -36 + 96 - 63
\]
\[
= -3
\]

The completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & -99 \\
0 & -63 \\
1 & -35 \\
2 & -15 \\
3 & -3 \\
\end{array}
\]

---

3. \( h(x) = -7x^2 + 42x - 65 \)

The table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & \\
0 & \\
1 & \\
2 & \\
3 & \\
\end{array}
\]

#### Substituting \( x = -1 \):
\[
h(-1) = -7(-1)^2 + 42(-1) - 65
\]
\[
= -7(1) + 42(-1) - 65
\]
\[
= -7 - 42 - 65
\]
\[
= -114
\]

#### Substituting \( x = 0 \):
\[
h(0) = -7(0)^2 + 42(0) - 65
\]
\[
= -65
\]

#### Substituting \( x = 1 \):
\[
h(1) = -7(1)^2 + 42(1) - 65
\]
\[
= -7(1) + 42(1) - 65
\]
\[
= -7 + 42 - 65
\]
\[
= -30
\]

#### Substituting \( x = 2 \):
\[
h(2) = -7(2)^2 + 42(2) - 65
\]
\[
= -7(4) + 42(2) - 65
\]
\[
= -28 + 84 - 65
\]
\[
= -9
\]

#### Substituting \( x = 3 \):
\[
h(3) = -7(3)^2 + 42(3) - 65
\]
\[
= -7(9) + 42(3) - 65
\]
\[
= -63 + 126 - 65
\]
\[
= -2
\]

The completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & -114 \\
0 & -65 \\
1 & -30 \\
2 & -9 \\
3 & -2 \\
\end{array}
\]

---

4. \( k(x) = -5x^2 + 20x - 1 \)

The table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & \\
0 & \\
1 & \\
2 & \\
3 & \\
\end{array}
\]

#### Substituting \( x = -1 \):
\[
k(-1) = -5(-1)^2 + 20(-1) - 1
\]
\[
= -5(1) + 20(-1) - 1
\]
\[
= -5 - 20 - 1
\]
\[
= -26
\]

#### Substituting \( x = 0 \):
\[
k(0) = -5(0)^2 + 20(0) - 1
\]
\[
= -1
\]

#### Substituting \( x = 1 \):
\[
k(1) = -5(1)^2 + 20(1) - 1
\]
\[
= -5(1) + 20(1) - 1
\]
\[
= -5 + 20 - 1
\]
\[
= 14
\]

#### Substituting \( x = 2 \):
\[
k(2) = -5(2)^2 + 20(2) - 1
\]
\[
= -5(4) + 20(2) - 1
\]
\[
= -20 + 40 - 1
\]
\[
= 19
\]

#### Substituting \( x = 3 \):
\[
k(3) = -5(3)^2 + 20(3) - 1
\]
\[
= -5(9) + 20(3) - 1
\]
\[
= -45 + 60 - 1
\]
\[
= 14
\]

The completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & -26 \\
0 & -1 \\
1 & 14 \\
2 & 19 \\
3 & 14 \\
\end{array}
\]

---

5. \( m(x) = -4x^2 + 24x + 7 \)

The table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & \\
0 & \\
1 & \\
2 & \\
3 & \\
\end{array}
\]

#### Substituting \( x = -1 \):
\[
m(-1) = -4(-1)^2 + 24(-1) + 7
\]
\[
= -4(1) + 24(-1) + 7
\]
\[
= -4 - 24 + 7
\]
\[
= -21
\]

#### Substituting \( x = 0 \):
\[
m(0) = -4(0)^2 + 24(0) + 7
\]
\[
= 7
\]

#### Substituting \( x = 1 \):
\[
m(1) = -4(1)^2 + 24(1) + 7
\]
\[
= -4(1) + 24(1) + 7
\]
\[
= -4 + 24 + 7
\]
\[
= 27
\]

#### Substituting \( x = 2 \):
\[
m(2) = -4(2)^2 + 24(2) + 7
\]
\[
= -4(4) + 24(2) + 7
\]
\[
= -16 + 48 + 7
\]
\[
= 39
\]

#### Substituting \( x = 3 \):
\[
m(3) = -4(3)^2 + 24(3) + 7
\]
\[
= -4(9) + 24(3) + 7
\]
\[
= -36 + 72 + 7
\]
\[
= 43
\]

The completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & -21 \\
0 & 7 \\
1 & 27 \\
2 & 39 \\
3 & 43 \\
\end{array}
\]

---

6. \( n(x) = -8x^2 + 16x - 9 \)

The table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & \\
0 & \\
1 & \\
2 & \\
3 & \\
\end{array}
\]

#### Substituting \( x = -1 \):
\[
n(-1) = -8(-1)^2 + 16(-1) - 9
\]
\[
= -8(1) + 16(-1) - 9
\]
\[
= -8 - 16 - 9
\]
\[
= -33
\]

#### Substituting \( x = 0 \):
\[
n(0) = -8(0)^2 + 16(0) - 9
\]
\[
= -9
\]

#### Substituting \( x = 1 \):
\[
n(1) = -8(1)^2 + 16(1) - 9
\]
\[
= -8(1) + 16(1) - 9
\]
\[
= -8 + 16 - 9
\]
\[
= -1
\]

#### Substituting \( x = 2 \):
\[
n(2) = -8(2)^2 + 16(2) - 9
\]
\[
= -8(4) + 16(2) - 9
\]
\[
= -32 + 32 - 9
\]
\[
= -9
\]

#### Substituting \( x = 3 \):
\[
n(3) = -8(3)^2 + 16(3) - 9
\]
\[
= -8(9) + 16(3) - 9
\]
\[
= -72 + 48 - 9
\]
\[
= -33
\]

The completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & -33 \\
0 & -9 \\
1 & -1 \\
2 & -9 \\
3 & -33 \\
\end{array}
\]

---

7. \( p(x) = -9x^2 + 72x + 16 \)

The table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & \\
0 & \\
1 & \\
2 & \\
3 & \\
\end{array}
\]

#### Substituting \( x = -1 \):
\[
p(-1) = -9(-1)^2 + 72(-1) + 16
\]
\[
= -9(1) + 72(-1) + 16
\]
\[
= -9 - 72 + 16
\]
\[
= -65
\]

#### Substituting \( x = 0 \):
\[
p(0) = -9(0)^2 + 72(0) + 16
\]
\[
= 16
\]

#### Substituting \( x = 1 \):
\[
p(1) = -9(1)^2 + 72(1) + 16
\]
\[
= -9(1) + 72(1) + 16
\]
\[
= -9 + 72 + 16
\]
\[
= 79
\]

#### Substituting \( x = 2 \):
\[
p(2) = -9(2)^2 + 72(2) + 16
\]
\[
= -9(4) + 72(2) + 16
\]
\[
= -36 + 144 + 16
\]
\[
= 124
\]

#### Substituting \( x = 3 \):
\[
p(3) = -9(3)^2 + 72(3) + 16
\]
\[
= -9(9) + 72(3) + 16
\]
\[
= -81 + 216 + 16
\]
\[
= 141
\]

The completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & -65 \\
0 & 16 \\
1 & 79 \\
2 & 124 \\
3 & 141 \\
\end{array}
\]

---

8. \( q(x) = -x^2 + 2x \)

The table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-2 & \\
-1 & \\
0 & \\
1 & \\
2 & \\
\end{array}
\]

#### Substituting \( x = -2 \):
\[
q(-2) = -(-2)^2 + 2(-2)
\]
\[
= -(4) + 2(-2)
\]
\[
= -4 - 4
\]
\[
= -8
\]

#### Substituting \( x = -1 \):
\[
q(-1) = -(-1)^2 + 2(-1)
\]
\[
= -(1) + 2(-1)
\]
\[
= -1 - 2
\]
\[
= -3
\]

#### Substituting \( x = 0 \):
\[
q(0) = -(0)^2 + 2(0)
\]
\[
= 0
\]

#### Substituting \( x = 1 \):
\[
q(1) = -(1)^2 + 2(1)
\]
\[
= -(1) + 2(1)
\]
\[
= -1 + 2
\]
\[
= 1
\]

#### Substituting \( x = 2 \):
\[
q(2) = -(2)^2 + 2(2)
\]
\[
= -(4) + 2(2)
\]
\[
= -4 + 4
\]
\[
= 0
\]

The completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-2 & -8 \\
-1 & -3 \\
0 & 0 \\
1 & 1 \\
2 & 0 \\
\end{array}
\]

---

9. \( r(x) = -3x^2 + 6x + 15 \)

The table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & \\
0 & \\
1 & \\
2 & \\
3 & \\
\end{array}
\]

#### Substituting \( x = -1 \):
\[
r(-1) = -3(-1)^2 + 6(-1) + 15
\]
\[
= -3(1) + 6(-1) + 15
\]
\[
= -3 - 6 + 15
\]
\[
= 6
\]

#### Substituting \( x = 0 \):
\[
r(0) = -3(0)^2 + 6(0) + 15
\]
\[
= 15
\]

#### Substituting \( x = 1 \):
\[
r(1) = -3(1)^2 + 6(1) + 15
\]
\[
= -3(1) + 6(1) + 15
\]
\[
= -3 + 6 + 15
\]
\[
= 18
\]

#### Substituting \( x = 2 \):
\[
r(2) = -3(2)^2 + 6(2) + 15
\]
\[
= -3(4) + 6(2) + 15
\]
\[
= -12 + 12 + 15
\]
\[
= 15
\]

#### Substituting \( x = 3 \):
\[
r(3) = -3(3)^2 + 6(3) + 15
\]
\[
= -3(9) + 6(3) + 15
\]
\[
= -27 + 18 + 15
\]
\[
= 6
\]

The completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-1 & 6 \\
0 & 15 \\
1 & 18 \\
2 & 15 \\
3 & 6 \\
\end{array}
\]

---

10. \( s(x) = -12x^2 - 20 \)

The table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-2 & \\
-1 & \\
0 & \\
1 & \\
2 & \\
\end{array}
\]

#### Substituting \( x = -2 \):
\[
s(-2) = -12(-2)^2 - 20
\]
\[
= -12(4) - 20
\]
\[
= -48 - 20
\]
\[
= -68
\]

#### Substituting \( x = -1 \):
\[
s(-1) = -12(-1)^2 - 20
\]
\[
= -12(1) - 20
\]
\[
= -12 - 20
\]
\[
= -32
\]

#### Substituting \( x = 0 \):
\[
s(0) = -12(0)^2 - 20
\]
\[
= -20
\]

#### Substituting \( x = 1 \):
\[
s(1) = -12(1)^2 - 20
\]
\[
= -12(1) - 20
\]
\[
= -12 - 20
\]
\[
= -32
\]

#### Substituting \( x = 2 \):
\[
s(2) = -12(2)^2 - 20
\]
\[
= -12(4) - 20
\]
\[
= -48 - 20
\]
\[
= -68
\]

The completed table is:
\[
\begin{array}{c|c}
x & y \\
\hline
-2 & -68 \\
-1 & -32 \\
0 & -20 \\
1 & -32 \\
2 & -68 \\
\end{array}
\]

---

Final Answer

\[
\boxed{
\begin{array}{c|c}
\text{Function} & \text{Completed Table} \\
\hline
1. f(x) & \begin{array}{c|c}
x & y \\
\hline
-2 & -43 \\
-1 & -28 \\
0 & -19 \\
1 & -16 \\
2 & -19 \\
\end{array} \\
2. g(x) & \begin{array}{c|c}
x & y \\
\hline
-1 & -99 \\
0 & -63 \\
1 & -35 \\
2 & -15 \\
3 & -3 \\
\end{array} \\
3. h(x) & \begin{array}{c|c}
x & y \\
\hline
-1 & -114 \\
0 & -65 \\
1 & -30 \\
2 & -9 \\
3 & -2 \\
\end{array} \\
4. k(x) & \begin{array}{c|c}
x & y \\
\hline
-1 & -26 \\
0 & -1 \\
1 & 14 \\
2 & 19 \\
3 & 14 \\
\end{array} \\
5. m(x) & \begin{array}{c|c}
x & y \\
\hline
-1 & -21 \\
0 & 7 \\
1 & 27 \\
2 & 39 \\
3 & 43 \\
\end{array} \\
6. n(x) & \begin{array}{c|c}
x & y \\
\hline
-1 & -33 \\
0 & -9 \\
1 & -1 \\
2 & -9 \\
3 & -33 \\
\end{array} \\
7. p(x) & \begin{array}{c|c}
x & y \\
\hline
-1 & -65 \\
0 & 16 \\
1 & 79 \\
2 & 124 \\
3 & 141 \\
\end{array} \\
8. q(x) & \begin{array}{c|c}
x & y \\
\hline
-2 & -8 \\
-1 & -3 \\
0 & 0 \\
1 & 1 \\
2 & 0 \\
\end{array} \\
9. r(x) & \begin{array}{c|c}
x & y \\
\hline
-1 & 6 \\
0 & 15 \\
1 & 18 \\
2 & 15 \\
3 & 6 \\
\end{array} \\
10. s(x) & \begin{array}{c|c}
x & y \\
\hline
-2 & -68 \\
-1 & -32 \\
0 & -20 \\
1 & -32 \\
2 & -68 \\
\end{array} \\
\end{array}
}
\]