College Algebra

m = y1 - y0
x1 - x0

line with slope m containing the point (x0; y0) is
the equation y = m(x - x0) + y0.

line with slope m and y-intercept (0; b) is the
equation y = mx + b:

is a function of the form
f(x) = mx + b;
where m and b are real numbers with m ≠ 0. The domain of a linear function is (-¥,¥).

is a function of the form
f(x) = b;
where b is real number. The domain of a constant function is (- ¥,¥). This is basically a seperate classification for when m = 0

∆f
∆x
=
f(b) - f(a)
b - a
the average rate of change over [a; b] is the slope of
the line which connects (a; f(a)) and (b; f(b)). This is called the secant line

Suppose a, b, c, h, and k are
real numbers with a ≠ 0.
• If f(x) = a(x - h)2 + k, the vertex of the graph of y = f(x) is the point (h, k).
• If f(x) = ax2 + bx + c, the vertex of the graph of y = f(x) is the point 

f(x) = a(x - h)2 + k, a ≠ 0

f(x) = ax2 + bx + c, a ≠ 0

is the quantity b2 - 4ac

• If b2 - 4ac < 0, the equation ax2 + bx + c = 0 has no real solutions.
• If b2 - 4ac = 0, the equation ax2 + bx + c = 0 has exactly one real solution.
• If b2 - 4ac > 0, the equation ax2 + bx + c = 0 has exactly two real solutions.

1. Rewrite the inequality, if necessary, as a quadratic function f(x) on one side of the inequality
and 0 on the other.
2. Find the zeros of f and place them on the number line with the number 0 above them.
3. Choose a real number, called a test value, in each of the intervals determined in step 2.
4. Determine the sign of f(x) for each test value in step 3, and write that sign above the
corresponding interval.
5. Choose the intervals which correspond to the correct sign to solve the inequality.