What is the vertex of x^2 -10x + 21
1 answer:
Answer:
(5, - 4)
Step-by-step explanation:
Given a quadratic in standard form : ax² + bx + c : a ≠ 0
Then the x- coordinate of the vertex is
= -
x² - 10x + 21 ← is in standard form
with a = 1, b = - 10, so
= -
= 5
Substitute x = 5 into the quadratic for the corresponding value of y
y = 5² - 10(5) + 21 = 25 - 50 + 21 = - 4
vertex = (5, - 4)
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Answer:
The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.
Step-by-step explanation:
Given : The position of an object along a vertical line is given by , where s is measured in feet and t is measured in seconds.
To find : What is the maximum velocity of the object in the time interval [0, 4]?
Solution :
The velocity is rate of change of distance w.r.t time.
Distance in terms of t is given by,
Derivate w.r.t. time,
It is a quadratic function so its maximum is at vertex of the function.
The x point of the function is given by,
Where, a=-3, b=6 and c=7
As 1 lie between interval [0,4]
Substitute t=1 in the function,
Th maximum velocity is 10 ft/s.
Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.
Answer:
z = 3
Step-by-step explanation:
Since the points are collinear then the slopes between the points are equal.
Calculate the slope m using the slope formula
m =
with (x₁, y₁ ) = P (2, - 3) and (x₂, y₂ ) = Q (3, - 2)
m = = 1
Repeat with
(x₁, y₁ ) = Q (3, - 2) and (x₂, y₂ ) = R (8, z )
m = =
, then
= 1 ( multiply both sides by 5 )
z + 2 = 5 ( subtract 2 from both sides )
z = 3