(a) By the fundamental theorem of calculus,
<em>v(t)</em> = <em>v(0)</em> + ∫₀ᵗ <em>a(u)</em> d<em>u</em>
The particle starts at rest, so <em>v(0)</em> = 0. Computing the integral gives
<em>v(t)</em> = [2/3 <em>u</em> ³ + 2<em>u</em> ²]₀ᵗ = 2/3 <em>t</em> ³ + 2<em>t</em> ²
(b) Use the FTC again, but this time you want the distance, which means you need to integrate the <u>speed</u> of the particle, i.e. the absolute value of <em>v(t)</em>. Fortunately, for <em>t</em> ≥ 0, we have <em>v(t)</em> ≥ 0 and |<em>v(t)</em> | = <em>v(t)</em>, so speed is governed by the same function. Taking the starting point to be the origin, after 8 seconds the particle travels a distance of
∫₀⁸ <em>v(u)</em> d<em>u</em> = ∫₀⁸ (2/3 <em>u</em> ³ + 2<em>u</em> ²) d<em>u</em> = [1/6 <em>u</em> ⁴ + 2/3 <em>u</em> ³]₀⁸ = 1024
Answer:
(x - 1)(x - 1)
Step-by-step explanation:
x² - 2x + 1
x² - x - x + 1
x(x - 1) - (x - 1)
(x - 1)(x - 1)
(x - 1)²
Answer:
its A because you go 3 to the right and you go 1 down because its a negative
Answer:
D. (3. —1)
Step-by-step explanation:
Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
It's vertex is the point 
In which
Where
If a<0, the vertex is a maximum point, that is, the maximum value happens at 
, and it's value is 
.
Turning point of a quadratic function:
The turning point of a quadratic function is the vertex.
y = x2 — 6x + 8
This means that 
Now, we find the vertex.
(x,y) = (3,-1), and the correct answer is given by option D.
