Homework-3 college algebra

Question

Suppose the​ velocity, v, of an object is given by v=2t²+3t−5, where t is time in seconds.​ (Here t can be positive or​ negative.) Find the intervals where the velocity is negative.

Answer 1

v(t) = 2t^2 +3t -5= 0

factor

(2t +5)(t-1) = 0

t= 1, -2.5

the interval (-2.5, 1) or -2.5<t<1 is where velocity is negative

or graph the polynomial, plots some points or use a graphing calculator

it crosses the x axis at t=1 and -2.5

it's an upward opening parabola which is below the x axis betwen those two x intercepts

plot t on the x axis and velocity on the y axis.

Answer 2

When a function crosses from being positive to negative, and vice versa, it has to go through 0. (This is a special case of the Intermediate Value Theorem.) Therefore, you should start by setting v=0 and finding the solutions to that equation. For this equation, you will find two solutions, which will split the number line into three segments. From there, you can plug in a value within those intervals to find out if it is positive or negative.