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 (-∞, ∞)
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This is t for which:
So
They will be at the same position after 2 hours.