Y = mx+b (slope intercept form). I imagine the point is (2,2) and m = 0. So the answer is y = 2. A sketch would be a horizontal line on the x-y plane going through (2,2).
The area between the two functions is 0
<h3>How to determine the area?</h3>
The functions are given as:
f₁(x)= 1
f₂(x) = |x - 2|
x ∈ [0, 4]
The area between the functions is
A = ∫[f₂(x) - f₁(x) ] dx
The above integral becomes
A = ∫|x - 2| - 1 dx (0 to 4)
When the above is integrated, we have:
A = [(|x - 2|(x - 2))/2 - x] (0 to 4)
Expand the above integral
A = [(|4 - 2|(4 - 2))/2 - 4] - [(|0 - 2|(0 - 2))/2 - 0]
This gives
A = [2 - 4] - [-2- 0]
Evaluate the expression
A = 0
Hence, the area between the two functions is 0
Read more about areas at:
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Taxable income = 3854
they have been taking taxes out in the amount of 344 for 12 months.....344* 12 = 4128....so they took out too much....so there is a refund of 4128 - 3854 =
274 refund
The vertex (minimum) of the quadratic ax² +bx +c is located at x=-b/(2a). This means the minimum value of f(x) will be found at x = -3/(2*1) = -1.5.
Since the vertex of the quadratic is less than 0, the maximum value of the quadratic will be found at x=2, the end of the interval farthest from the vertex.
On the given interval, ...
the absolute minimum value of f is f(-1.5) = ln(1.75) ≈ 0.559616
the absolute maximum value of f is f(2) = ln(14) ≈ 2.639057