Over which interval is the graph of f(x) = x2 + 5x + 6 increasing?
Answer:
Step-by-step explanation:
Given is a function
\(f(x) = x^2 + 5x + 6\)
Since leading term is positive,
the parabola is open up.
Let us write this in vertex form after completing the square
\(f(x) = x^2 +2(\frac{5}{2})x+\frac{25}{4} -\frac{25}{4}+6\\=(x+(\frac{5}{2})^2-\frac{1}{4}\)
Hence we get vertex= (-5/2,-1/4)
Since this is a parabola open up we have the minimum point is only one at x=-5/2
So we have the function decreasing for x<-2.5 and increasing for x>-2.5
Decreases in \((-\infty, -2.5)\)
Increases in \((-2.5,\infty)\)
Answer:
(–5, ∞)
Step-by-step explanation:
This is a vertical parabola open upward
The vertex represent a minimum
The vertex is the point (-5,-6.5)
The domain is all real numbers
The range is the interval [-6.5,∞)
so
At the left of the x-coordinate of the vertex the function is decreasing and at the right of the x-coordinate of the vertex the function is increasing
therefore
The function is increasing in the interval (-5,∞) and the function is decreasing in the interval (-∞,-5)