Answer:
y = -2x^2 - 4x - 1
Step-by-step explanation:
We can see that the graph passes through (-2, -1), (-1, 1) and (0, -1).
Let's solve
ax^2 + bx + c = y
a(-2)^2 + b(-2) + c = -1
4a - 2b + c = -1
a(-1)^2 + b(-1) + c = 1
a - b + c = 1
a0^2 + b0 + c = -1
c = -1
we got c = -1 so we input it into the other 2
4a - 2b - 1 = -1
4a - 2b = 0
2a - b = 0
2a = b
a - b - 1 = 1
a - b = 2
a = b + 2
Let's input b = 2a
a = 2a + 2
-a = 2
a = -2
b = 2a = 2*(-2) = -4
c = -1
y = -2x^2 - 4x - 1
Answer: I thin it's A because you just multiply 19*6
Step-by-step explanation:
(6,8) is the answer also this could be the answer too (6,2)
hope this helps
The maximum height the ball achieves before landing is 682.276 meters at t = 0.
<h3>What are maxima and minima?</h3>
Maxima and minima of a function are the extreme within the range, in other words, the maximum value of a function at a certain point is called maxima and the minimum value of a function at a certain point is called minima.
We have a function:
h(t) = -4.9t² + 682.276
Which represents the ball's height h at time t seconds.
To find the maximum height first find the first derivative of the function and equate it to zero
h'(t) = -9.8t = 0
t = 0
Find second derivative:
h''(t) = -9.8
At t = 0; h''(0) < 0 which means at t = 0 the function will be maximum.
Maximum height at t = 0:
h(0) = 682.276 meters
Thus, the maximum height the ball achieves before landing is 682.276 meters at t = 0.
Learn more about the maxima and minima here:
brainly.com/question/6422517
#SPJ1
