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aliina [53]
1 year ago
6

The following table gives a partial set of values of a polynomial function. x −2 −1 0 1 2 3 g(x) 14 2 0 4 6 −2 Between which two

values would a relative minimum most likely occur? (2 points) −2 and 0 0 and 2 −1 and 1 1 and 3
Mathematics
1 answer:
motikmotik1 year ago
3 0

From -1 to 0, the function goes decreases, and from 0 to 1 the function goes increases. Then we will have the minimum between -1 and 1.

<h3>Between which two values would a relative minimum most likely occur?</h3>

Here we have the table:

  • x: -2, -1, 0, 1, 2, 3
  • g(x): 14, 2, 0, 4, 6, -2

You can see that:

g(-1) = 2

g(0) = 0

So the function goes downwards from x = -1 to x = 0.

g(1) = 4

So the function goes upwards between 0 and 4.

Then, from -1 to 0, the function goes decreases, and from 0 to 1 the function goes increases. Then we will have the minimum between -1 and 1.

If you want to learn more about polynomial functions:

brainly.com/question/7693326

#SPJ1

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A ball is thrown up in the air, and it's height (in feet) as a function of time (in seconds) can be written as h(t) = -16t2 + 32
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Answer:

1) Using properties of the quadratic equation:

Here the height equation is:

h(t) = -16t^2 + 32t + 6

We can see that the leading coefficient is negative, this means that the arms of the graph will open downwards.

Then, the vertex of the quadratic equation will be the maximum.

Remember that for a general quadratic equation:

a*x^2 + b*x + c = y

the x-value of the vertex is:

x = -b/2a

Then in our case, the vertex is at:

t = -32/(2*-16) = 1

The maximum height will be the height equation evaluated in this time:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

2) Second method, using physics.

We know that an object reaches its maximum height when the velocity is equal to zero (the velocity equal to zero means that, at this point, the object stops going upwards).

If the height equation is:

h(t) = -16*t^2 + 32*t + 6

the velocity equation is the first derivation of h(t)

Remember that for a function f(x) = a*x^n

we have that:

df(x)/dx = n*a*x^(n-1)

Then:

v(t) = dh(t)/dt = 2*(-16)*t + 32 + 0

v(t) = -32*t + 32

Now we need to find the value of t such that the velocity is equal to zero:

v(t) = 0 = -32*t + 32

       32*t = 32

            t = 32/32 = 1

So the maximum height is at t = 1

(same as before)

Now we just need to evaluate the height equation in t = 1:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

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3 years ago
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