Respuesta :

Part A

Answer:  -4

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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

Answer: 4 units

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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

Answer: m(x) never exceeds h(x)

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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.

Ver imagen jimthompson5910
sqdancefan

Answer:

  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).

Ver imagen sqdancefan

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