Determine the domain of the function f(x)=1/3x^2+10x+8
1 answer:
The function is
f(x) = (1/3)x² + 10x + 8
Write the function in standard form for a parabola.
f(x) = (1/3)[x² + 30x] + 8
= (1/3)[ (x+15)²- 225] + 8
= (1/3)(x+15)² -75 + 8
f(x) = (1/3)(x+15)² - 67
This is a parabola with vertex at (-15, -67).
The axis of symmetry is x = -15
The curve opens upward because the coefficient of x² is positive.
As x -> - ∞, f -> +∞.
As x -> +∞, f -> +∞
The domain is all real values of x (see the graph below).
Answer: The domain is (-∞, ∞)
f(x) = (1/3)x² + 10x + 8
Write the function in standard form for a parabola.
f(x) = (1/3)[x² + 30x] + 8
= (1/3)[ (x+15)²- 225] + 8
= (1/3)(x+15)² -75 + 8
f(x) = (1/3)(x+15)² - 67
This is a parabola with vertex at (-15, -67).
The axis of symmetry is x = -15
The curve opens upward because the coefficient of x² is positive.
As x -> - ∞, f -> +∞.
As x -> +∞, f -> +∞
The domain is all real values of x (see the graph below).
Answer: The domain is (-∞, ∞)
You might be interested in
Answer:
Step-by-step explanation:
From the given information,
The ten observation data on october snow cover for Eurasia during the years is 6.5, 12.0, 14.9, 10.0, 10.7, 7.9, 21.9, 12.5, 14.5, 9.2
What would you report as a representative, or typical, value of October snow cover for this period, and what prompted your choice?
For the given data, 21.9 is an outlier, so trimmed mean would be good choice for the researcher,
Remove the smallest and the largest values to compute the trimmed mean