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nordsb [41]
3 years ago
7

On a coordinate plane, a parabola opens up. It goes through (negative 8, negative 2), has a vertex at (negative 5, negative 6.5)

, goes through (negative 2, negative 2), and has a y-intercept at (0, 6). Over which interval is the graph of f(x) = one-halfx2 + 5x + 6 increasing? (–6.5, ∞) (–5, ∞) (–∞, –5) (–∞, –6.5)
Mathematics
2 answers:
dimaraw [331]3 years ago
7 0

Answer:

The function is increasing in the interval (-5,∞)

Step-by-step explanation:

we have

f(x)=\frac{1}{2}x^2+5x+6

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)

erastovalidia [21]3 years ago
4 0

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)

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