IM BEGGING PLEASE ANSWER THIS MATH PROBLEM

Part A

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Work Shown:

Plug x = 4 into h(x)

h(x) = (1/2)*(x-2)^2

h(4) = (1/2)*(4-2)^2

h(4) = (1/2)*(2)^2

h(4) = (1/2)*4

h(4) = 2

We'll use this value later.

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The points shown in the m(x) table all fall on the same straight line. Pick the first two points (8,2) and (10,3) to find the slope to be...

m = (y2 - y1)/(x2 - x1)

m = (3 - 2)/(10 - 8)

m = 1/2

Plug m = 1/2 and (x1,y1) = (8,2) into the point slope formula. Solve for y.

y - y1 = m(x - x1)

y - 2 = (1/2)(x - 8)

y - 2 = (1/2)x + (1/2)(-8)

y - 2 = (1/2)*x - 4

y - 2 + 2 = (1/2)*x - 4 + 2

y = (1/2)*x - 2

Therefore m(x) = (1/2)x - 2

Now plug in x = 16

m(x) = (1/2)x - 2

m(16) = (1/2)(16) - 2

m(16) = 8 - 2

m(16) = 6

The point (16,6) is on the m(x) line.

In the diagram below, the point (16,6) is represented by point E.

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Finally,

h(4) - m(16) = 2 - 6

h(4) - m(16) = -4

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Part B

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Work Shown:

Plug x = 0 into h(x) to find the y intercept for function h

h(x) = (1/2)*(x-2)^2

h(0) = (1/2)*(0-2)^2

h(0) = (1/2)*(-2)^2

h(0) = (1/2)*4

h(0) = 2

The y intercept here is 2.

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Plug x = 0 into the m(x) function found back in part A

m(x) = (1/2)x - 2

m(0) = (1/2)(0) - 2

m(0) = -2

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Subtract the y intercepts and use absolute value to make sure the result is positive

|2 - (-2)| = |2 + 2| = |4| = 4

The distance between the two y intercepts is 4 units.

Check out the diagram below. Note the distance between points G and F (which are the y intercepts for m(x) and h(x) respectively) is exactly 4 units.

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Part C

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Work Shown:

If you compare the two graphs (see diagram below) you'll notice that the parabola is always above the diagonal line. Therefore, m(x) is never larger than h(x) for any x input.

  a) -4

  b) 4 units

  c) no values; h(x) is always greater than m(x)

Step-by-step explanation:

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a) In order to evaluate the difference of the functions, we need to know the value of m(16). We can tell that the x-values in the table increase by 2 at the same time the m(x) values increase by 1. Extending the table in this fashion to x=16, we add the column x=16, m(x)=6. Thus m(16) = 6.

Evaluating the function h(x) for x=4, we get ...

  h(4) = (1/2)(4 -2)^2 = (1/2)(4)

  h(4) = 2

So, the desired difference is ...

  h(4) -m(16) = 2 -6 = -4

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b) The y-intercept of h(x) is the value of h(0):

  h(0) = 1/2(0 -2)^2 = 2

We already know that m(x) increases by 1 when x increases by 2, so the slope is 1/2. Using a point-slope form of the equation of a line, we can write m(x) as ...

  y = m(x -a) +b . . . . . . for a line with slope m through the point (a, b)

  m(x) = (1/2)(x -8) +2 = 1/2x -2

Then the y-intercept of m(x) is -2.

The two y-intercepts differ in value by ...

  2 -(-2) = 4

The y-intercepts are 4 units apart.

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c) A graph of the two functions shows that h(x) is more than m(x) for all values of x. We can show that analytically by looking at the difference m(x) -h(x).

Subtracting h(x) from m(x), we want to know where this difference is positive.

  m(x) -h(x) = (1/2x -2) -(1/2)(x -2)^2

  = 1/2x -2 -1/2(x^2 -4x +4)

  = -1/2x^2 +5/2x -4

  = -1/2(x^2 -5x +8)

We can rewrite the quadratic factor in vertex form as ...

  x^2 -5x +8 = (x^2 -5x +6.25) +1.75 = (x -2.5)^2 +1.75

This factor is positive everywhere, so the difference

  m(x) -h(x) = -1/2(some positive number)

will be negative everywhere. That is, there are no values of x that make m(x) greater than h(x).

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step-by-step explanation:

both the histogram and the stem and leaf plot will visually show how many shoppers were of various ages within each range of data.   the histogram does not give us the specific ages of these shoppers unlike the stem and leaf plot.a histogram does not give us a true range of the data because we do not know exactly how old the shoppers were. a stem and leaf plot will give exact ages, so we can determine the exact range along with all measures of center and the others measures of center.   with the histogram, we cannot find any of these.

step-by-step explanation: