How do you do this problem?
2 answers:
Answer:
x = -0.5
Step-by-step explanation:
I have marked on the graph the values of ƒ(x) that correspond to the different x-values.
At x = -2.5, ƒ(x) is decreasing, concave up, and > 0.
At x = -1.5, ƒ(x) is decreasing, concave down, and < 0.
At x = -0.5, ƒ(x) is increasing, concave up, and < 0.
At x = 0.5, ƒ(x) is increasing, concave up, and > 0.
At x = 2.5, ƒ(x) is decreasing, concave up, and > 0.
The only point for which ƒ(x) is simultaneously increasing, concave up, and less than zero is x = -0.5.
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Answer:
Step-by-step explanation:
Given the following quadratic equation:
The steps to solve it are:
1. Move the terms to one side of the equation:
2. Apply the Quadratic formula .
In this case we can identify that:
Then, substituting these values into the Quadratic formula we get the following solutions: