HELP ME!!!
1 answer:
Answer:
-2, 8/3
Step-by-step explanation:
You can consider the area to be that of a trapezoid with parallel bases f(a) and f(4), and width (4-a). The area of that trapezoid is ...
A = (1/2)(f(a) +f(4))(4 -a)
= (1/2)((3a -1) +(3·4 -1))(4 -a)
= (1/2)(3a +10)(4 -a)
We want this area to be 12, so we can substitute that value for A and solve for "a".
12 = (1/2)(3a +10)(4 -a)
24 = (3a +10)(4 -a) = -3a² +2a +40
3a² -2a -16 = 0 . . . . . . subtract the right side
(3a -8)(a +2) = 0 . . . . . factor
Values of "a" that make these factors zero are ...
a = 8/3, a = -2
The values of "a" that make the area under the curve equal to 12 are -2 and 8/3.
_____
<em>Alternate solution</em>
The attachment shows a solution using the numerical integration function of a graphing calculator. The area under the curve of function f(x) on the interval [a, 4] is the integral of f(x) on that interval. Perhaps confusingly, we have called that area f(a). As we have seen above, the area is a quadratic function of "a". I find it convenient to use a calculator's functions to solve problems like this where possible.
You might be interested in
The equation of the first line can be written in point-slope form as
.. y = 3(x +1) -8
or
.. 3x -y = 5
The equation of the second line can be written in 2-point form as
.. y = (-1-3)/(10-(-6))*(x +6) +3
.. y = (-1/4)(x +6) +3
or
.. x +4y = 6
A graph shows the solution to this system is (x, y) = (2, 1).
_____
The second equation can be used to write an expression for x:
.. x = 6 -4y
This can be substituted into the first equation.
.. 3(6 -4y) -y = 5
.. 18 -13y = 5 . . . . . . . collect terms
.. 13 = 13y . . . . . . . . . add 13y-5
.. 1 = y . . . . . . . . . . . . divide by 13
From the above equation for x
.. x = 6 -4*1 = 2
.. y = 3(x +1) -8
or
.. 3x -y = 5
The equation of the second line can be written in 2-point form as
.. y = (-1-3)/(10-(-6))*(x +6) +3
.. y = (-1/4)(x +6) +3
or
.. x +4y = 6
A graph shows the solution to this system is (x, y) = (2, 1).
_____
The second equation can be used to write an expression for x:
.. x = 6 -4y
This can be substituted into the first equation.
.. 3(6 -4y) -y = 5
.. 18 -13y = 5 . . . . . . . collect terms
.. 13 = 13y . . . . . . . . . add 13y-5
.. 1 = y . . . . . . . . . . . . divide by 13
From the above equation for x
.. x = 6 -4*1 = 2