Mathematics, 10.03.2020 17:38 sickomode2048

The equation of function h is h(x) = ½ (x-2)^2. The table shows some of the values of function m.

X : 8 | 10 | 12 | 14
m(x) : 2 | 3 | 4 | 5

Part A : What is the value of h(4) - m(16). Show your work or explain your answer.

Part B : If both of these functions are graphed on the same coordinate plane, how far apart are the y-intercepts? Explain how you obtained your answer.

Part C : For what values of x does m(x) exceed h(x)? Explain your answer.

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Answer from: vic2nvsty

Part A: - 2Part B: the y-intercepts are 4 units apartPart C: m(x) never exceeds h(x)

Explanation:

1. Function h(x) (given)

$h(x)=\dfrac{1}{2}(x-2)^2$

2. Table with some values of the function m(x) (given)

x : 8 10 12 14

m(x): 2 3 4 5

Part A : What is the value of h(4) - m(16). Show your work or explain your answer.

To find m(16) you need to find the function m(x).

First, notice that it is a linear functio, because for constant increements of the input, x-values, there are constant increments of the output, m(x).

The point-slope equation of a linear function is:

y-y_1=m(x-x_1)

Where m is the slope and x₁ and y₁ are the pair of coordinates of any point that belongs to the line.

Calculate m:

$m=\dfrac{rise}{run}=\dfrac{\Delta y}{\Delta x}=\dfrac{y_2-y_1}{x_2-x_1}$

Use any two points. For instance (8,2), (10,3)

$m=\dfrac{3-2}{10-8}=\dfrac{1}{2}$

Replace m and any point into the point-slope equation:

$y-2=\dfrac{1}{2}(x-8)\\\\\\y=\dfrac{1}{2}x-4+2\\ \\ \\ y=\dfrac{1}{2}x-2$

$m(x)=\dfrac{1}{2}x-2$

Calculate h(4) and m(16)"

$h(4)=\dfrac{1}{2}(4-2)^2=\dfrac{1}{2}(2)^2=2$

$m(16)=\dfrac{1}{2}(16-2)=4$

$h(4)-m(16)=2-4=-4\\\\h(4)-m(16)=-2\leftarrow answer$

Part B : If both of these functions are graphed on the same coordinate plane, how far apart are the y-intercepts? Explain how you obtained your answer.

The y-intercepts are the values of y where the graph of the function crosses the y-axis.

That happens when x = 0.

Thus, to find the y-intercepts algebraically, you replace x with 0 in the equation of the function.

For h(x):

$h(0)=\dfrac{1}{2}(0-2)^2=\dfrac{1}{2}(-2)^2=2$

For m(x):

$m(0)=m(0)=\dfrac{1}{2}(0-2)=-2$

The distance is the absolute value of the difference:

|h(0)-m(0)|=|2-(-2)|=|2+2|=|4|=4

Part C : For what values of x does m(x) exceed h(x)?

Subtract the functions and in for what values of x the difference is positive (greater than 0)

$m(x)-h(x)0\\\\\\\dfrac{1}{2}x-2-\bigg( \dfrac{1}{2}(x-2)^2\bigg)\\ \\ \\ \dfrac{1}{2}x-2-\dfrac{1}{2}(x^2-2x+4)\\ \\ \\ \dfrac{1}{2}x-2-\dfrac{1}{2}x^2+x-2\\ \\ \\ -\dfrac{1}{2}x^2+\dfrac{3}{2}x-4$

Check when that is greater than 0:

$-\dfrac{1}{2}x^2+\dfrac{3}{2}x-40\\\\\\x^2-3x+8$

Find the discriminant of the quadratic function:

b^2-4ac=(-3)^2-4(1)(8)=9-32=-23

Since the discriminant is negative, that equation never crosses the x-axis.

Then, it will always have the same sign.

Substitute x = 0:

(0)^2-3(0)+8=8

Since 8 is greater than 0, you conclude that the left side is always greater than 0 and the inequality does not have solutions.

That means that m(x)never exceeds h(x).

Answer from: Quest

answer: about 22.67

step-by-step explanation:

Answer from: Quest

Ithink the answer is: it is likely.

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