Which explanation provides the best real-world scenario of the graph? If an object is dropped from a height of 38 feet, the func
1 answer:
The equation that models the movement of the object is:
Where,
t: time
a: acceleration due to gravity
v0: initial speed
h0: initial height
Suppose that the object falls with zero initial velocity and from a height of 38 feet.
The equation that models the problem is:
Answer:
If an object is dropped from a height of 38 feet, the function h (t) = -16t2 + 38 gives the height of the object after seconds
Where,
t: time
a: acceleration due to gravity
v0: initial speed
h0: initial height
Suppose that the object falls with zero initial velocity and from a height of 38 feet.
The equation that models the problem is:
Answer:
If an object is dropped from a height of 38 feet, the function h (t) = -16t2 + 38 gives the height of the object after seconds
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Answer:
There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.
Step-by-step explanation:
We notice that the expression on the left of the equation is a quadratic with leading term , which means that its graph is that of a parabola with branches going up.
Therefore, there can be three different situations:
1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.
2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.
3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.
So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.
We recall that the x-position of the vertex for a quadratic function of the form is given by the expression:
Since in our case and
, we get that the x-position of the vertex is:
Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:
This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.
We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:
See the graph produced in the attached image.
