Step-by-step explanation:
equation format: y = mx + b where m is slope and b is the y intercept.
From the points given we find that it is a horizontal line, which means slope or m = 0. So we know so far:
y = 0x + b or y = b
The y-intercept occurs where x = 0. This occurs at y = -7. So the equation for these two points:
EQUATION: y = -7
(Plotted graph attached)
The graph of Speedy's height in the air with time, which is based on a
quadratic function is a parabola.
Reasons:
The given parameters are;
The height of the cannon = 12 feet
Speedy's height after 2 seconds = 108 feet
Speedy's height after 3 seconds = 108 feet
Required: To select the best representation of the quadratic modelling Speedy's height, h(t), as a function of elapsed time, t?
Solution:
The path of Speedy's motion is a parabola
From the question, we have, that the initial height, at <em>t</em> = 0 is the height of the cannon = 12 feet
Therefore;
The equation has a constant term of 12
Given that the time it takes Speedy to rise above 108 feet and return to 108 feet = 3 - 2 = 1 second, we have;
The maximum height occurs between the 2nd and the 3rd second.
The path of a parabola is symmetric about the maximum point, therefore;
The maximum point occur at time
Therefore, the x-coordinate of the vertex is t = 2.5 s
From the general equation of a parabola, a·x² + b·x + c, the x-coordinate of the vertex is;
From the given option, we have the option; h(t) = -16·t² + 80·t + 12, which has;
Constant = 12
Vertex =
Therefore;
The best representation of Speedy's height is; <u>h(t) = -16·t² + 80·t + 12</u>
<em>The possible question options are;</em>
<em>h(t) = 1.07·t² + 5.33·t + 101.60</em>
<em>h(t) = 16·t² + 80·t + 12</em>
<em>h(t) = -1.07·t² + 5.33·t 101.60</em>
<em>h(t) = -16·t² + 80·t + 12 </em>
Learn more about projectile motion here: