10.
a. \(f(x) = 2x - 5\)
\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-2 & -9 \\
-1 & -7 \\
0 & -5 \\
1 & -3 \\
2 & -1 \\
\hline
\end{array}
\]
b. \(2x + y - 3 = -5\)
\[
\begin{array}{|c|c|c|}
\hline
x & y & (x, y) \\
\hline
-2 & 0 & (-2, 3) \\
-1 & 1 & (-1, 2) \\
0 & 2 & (0, 5) \\
1 & 3 & (1, 6) \\
2 & 4 & (2, 7) \\
\hline
\end{array}
\]
c. \(f(x) = 7-6x - x^2\)
\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-2 & -3 \\
-1 & 0 \\
0 & 5 \\
1 & 12 \\
2 & 21 \\
\hline
\end{array}
\]
d. \(f(x) = 1 - 4x\)
\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-2 & 9 \\
-1 & 5 \\
0 & 1 \\
1 & -3 \\
2 & -7 \\
\hline
\end{array}
\]
11.
a. The equation of the new curve is \(y = (x - 1)^2 - 4\).
b. To find where the curve intersects the x-axis, set \(y = 0\):
\((x - 1)^2 - 4 = 0\)
Solving this quadratic equation gives \(x = 3\) and \(x = -1\).
To find where the curve intersects the y-axis, set \(x = 0\):
\(y = (0 - 1)^2 - 4 = -3\).
c. The sketch represents a parabola opening upwards, centered at (1, -4), intersecting the x-axis at (3, 0) and (-1, 0), and the y-axis at (0, -3).
