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SpyIntel [72]
3 years ago
7

I need help on this pls

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
6 0

Answer:

Step-by-step explanation:

a). Given function is y = \frac{1}{2}x^3-2x+1

At x = -3,

y = \frac{1}{2}(-3)^3-2(-3)+1

y = -13.5 + 6 + 1

y = -6.5

For x = -2,

y = \frac{1}{2}(-2)^3-2(-2)+1

y = -4 + 4 + 1

y = 1

For x = 2

y = \frac{1}{2}(2)^2-2(2)+1

y = 4 - 4 + 1

y = 1

Therefore, table for the given function will be,

x       -3       -2       -1        0        1         2

y     -6.5       1      2.5      1       -0.5     1  

b). From the table given in part (a) we find the points lying on the graph are (-2, 1), (-1, 2.5) and (2, 1).

    These points justify the graph (C)

Therefore, graph (C) will be the answer.

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3. Classify each function according to whether it is a vertical stretch, a vertical compression, a horizontal
Vladimir79 [104]

The classifications of the functions are

  • A vertical stretch --- p(x) = 4f(x)
  • A vertical compression --- g(x) = 0.65f(x)
  • A horizontal stretch --- k(x) = f(0.5x)
  • A horizontal compression  --- h(x) = f(14x)

<h3>How to classify each function accordingly?</h3>

The categories of the functions are given as

  • A vertical stretch
  • A vertical compression
  • A horizontal stretch
  • A horizontal compression

The general rules of the above definitions are:

  • A vertical stretch --- g(x) = a f(x) if |a| > 1
  • A vertical compression --- g(x) = a f(x) if 0 < |a| < 1
  • A horizontal stretch --- g(x) = f(bx) if 0 < |b| < 1
  • A horizontal compression  --- g(x) = f(bx) if |b| > 1

Using the above rules and highlights, we have the classifications of the functions to be

  • A vertical stretch --- p(x) = 4f(x)
  • A vertical compression --- g(x) = 0.65f(x)
  • A horizontal stretch --- k(x) = f(0.5x)
  • A horizontal compression  --- h(x) = f(14x)

Read more about transformation at

brainly.com/question/1548871

#SPJ1

4 0
2 years ago
for a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height,
kolbaska11 [484]

Problem

For a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height, y, and time, x , when the projectile reaches the ground

Solution

We know that the x coordinate of a quadratic function is given by:

Vx= -b/2a

And the y coordinate correspond to the maximum value of y.

Then the best options are C and D but the best option is:

D) The maximum height is a y coordinate of the vertex of the quadratic function, which occurs when x = -b/2a

The projectile reaches the ground when the height is zero. The time when this occurs is the x-intercept of the zero of the function that is farthest to the right.​

7 0
1 year ago
Please help me with this last problem
sdas [7]
The answer to question #5 is 17
4 0
3 years ago
Read 2 more answers
Kenneth solved a certain number of problems and Harold solved 2 more than twice as many. Together they
Inessa [10]

H(arold) = 2 + 2*K(enneth)

H = 2 + 2K

H + K = 38

Substitution:

2 + 2K + K = 38

3K = 36

K = 12

Kenneth solved 12 problems and Harold solved (2(12) + 2 = 24 + 2 =) 26 problems

6 0
3 years ago
What is the m for this slope?
Blababa [14]

Answer:

-3/4x

Step-by-step explanation:

Down 3, right 4. We know it's negative because it's declining, and going down till you reach the x-coordinate of the point gives us -3, and going right 4 units to the location of the point.

7 0
3 years ago
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