How do you solve and write the following in interval notation: #x^2 - 4x - 21 ≤ 0#?

1 Answer
Jul 16, 2016

Answer:

#-3 <= x <= 7 #

Explanation:

#f(x) = x^2 - 4x - 21 <= 0#
First, solve f(x) = 0 to get the 2 x-intercepts (real roots) of f(x) = 0.
Find 2 numbers knowing sum (-b = 4) and product (c = -21). They are:
-3 and 7.
Plot these 2 end points (-3) and (7) on a number line. Since a > 0, the parabola opens upward. Between the real roots f(x) < 0, because a part of the parabola is below the x-axis.
Solution set by closed interval: [-3, 7]
The 2 end points (-3) and (7) are included in the solution set.
Graph:

------------------------- -3 ========0================= 7 -------------