Answer:
- Part B: the y-intercepts are 4 units apart
- Part C: m(x) never exceeds h(x)
Explanation:
1. Function h(x) (given)
[tex]h(x)=\dfrac{1}{2}(x-2)^2[/tex]
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:
[tex]y-y_1=m(x-x_1)[/tex]
Where m is the slope and x₁ and y₁ are the pair of coordinates of any point that belongs to the line.
Calculate m:
[tex]m=\dfrac{rise}{run}=\dfrac{\Delta y}{\Delta x}=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
Use any two points. For instance (8,2), (10,3)
[tex]m=\dfrac{3-2}{10-8}=\dfrac{1}{2}[/tex]
Replace m and any point into the point-slope equation:
[tex]y-2=\dfrac{1}{2}(x-8)\\\\\\y=\dfrac{1}{2}x-4+2\\ \\ \\ y=\dfrac{1}{2}x-2[/tex]
[tex]m(x)=\dfrac{1}{2}x-2[/tex]
Calculate h(4) and m(16)"
[tex]h(4)=\dfrac{1}{2}(4-2)^2=\dfrac{1}{2}(2)^2=2[/tex]
[tex]m(16)=\dfrac{1}{2}(16-2)=4[/tex]
[tex]h(4)-m(16)=2-4=-4\\\\h(4)-m(16)=-2\leftarrow answer[/tex]
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):
[tex]h(0)=\dfrac{1}{2}(0-2)^2=\dfrac{1}{2}(-2)^2=2[/tex]
For m(x):
[tex]m(0)=m(0)=\dfrac{1}{2}(0-2)=-2[/tex]
The distance is the absolute value of the difference:
[tex]|h(0)-m(0)|=|2-(-2)|=|2+2|=|4|=4[/tex]
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)
[tex]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[/tex]
Check when that is greater than 0:
[tex]-\dfrac{1}{2}x^2+\dfrac{3}{2}x-4>0\\\\\\x^2-3x+8<0[/tex]
Find the discriminant of the quadratic function:
[tex]b^2-4ac=(-3)^2-4(1)(8)=9-32=-23[/tex]
Since the discriminant is negative, that equation never crosses the x-axis.
Then, it will always have the same sign.
Substitute x = 0:
[tex](0)^2-3(0)+8=8[/tex]
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).
