Between y = 2x2 + 9x − 4 and y = −x2 + 6x + 2 for x in [−2, 2]
I assume you are looking for the area. A straightforward integration will give the algebraic area, since the curves cross at x=1, and the area changes sign.
∫[-2,2] (-x^2+6x+2)-(2x^2+9x-4) dx
= ∫[-2,2] -3x^2-3x+6 dx
= 8
However, to get the geometric area, you need to split the region into two parts:
∫[-2,1] (-x^2+6x+2)-(2x^2+9x-4) dx + ∫[1,2] (2x^2+9x-4)-(-x^2+6x+2) dx = 27/2 + 11/2 = 19
To compare the two functions y = 2x^2 + 9x - 4 and y = -x^2 + 6x + 2 for x in the interval [-2, 2], we need to evaluate the functions at the given x-values and compare their respective y-values.
Step 1: Find the y-values for the function y = 2x^2 + 9x - 4 for x in the interval [-2, 2].
Let's plug in the values of x in the given interval one by one and calculate the corresponding y-values.
For x = -2:
y = 2(-2)^2 + 9(-2) - 4
y = 2(4) - 18 - 4
y = 8 - 18 - 4
y = -14
For x = -1:
y = 2(-1)^2 + 9(-1) - 4
y = 2(1) - 9 - 4
y = 2 - 9 - 4
y = -11
For x = 0:
y = 2(0)^2 + 9(0) - 4
y = 0 + 0 - 4
y = -4
For x = 1:
y = 2(1)^2 + 9(1) - 4
y = 2 + 9 - 4
y = 7
For x = 2:
y = 2(2)^2 + 9(2) - 4
y = 2(4) + 18 - 4
y = 8 + 18 - 4
y = 22
So, for the function y = 2x^2 + 9x - 4, the corresponding y-values for x in the interval [-2, 2] are:
(-2, -14), (-1, -11), (0, -4), (1, 7), (2, 22).
Step 2: Find the y-values for the function y = -x^2 + 6x + 2 for x in the interval [-2, 2].
Similarly, let's evaluate the function at the given x-values.
For x = -2:
y = -(-2)^2 + 6(-2) + 2
y = -4 - 12 + 2
y = -14
For x = -1:
y = -(-1)^2 + 6(-1) + 2
y = -1 - 6 + 2
y = -5
For x = 0:
y = -(0)^2 + 6(0) + 2
y = 0 + 0 + 2
y = 2
For x = 1:
y = -(1)^2 + 6(1) + 2
y = -1 + 6 + 2
y = 7
For x = 2:
y = -(2)^2 + 6(2) + 2
y = -4 + 12 + 2
y = 10
So, for the function y = -x^2 + 6x + 2, the corresponding y-values for x in the interval [-2, 2] are:
(-2, -14), (-1, -5), (0, 2), (1, 7), (2, 10).
Now, we can compare the y-values for the two functions for each x-value in the interval [-2, 2]:
y-values for y = 2x^2 + 9x - 4: -14, -11, -4, 7, 22
y-values for y = -x^2 + 6x + 2: -14, -5, 2, 7, 10
Comparing the y-values, we can see that for each x-value in the interval [-2, 2], the y-values of the function y = 2x^2 + 9x - 4 are higher than the y-values of the function y = -x^2 + 6x + 2. Thus, y = 2x^2 + 9x - 4 is greater than y = -x^2 + 6x + 2 for x in the interval [-2, 2].
To find the values of y for the given equations within the range of x from -2 to 2, we can substitute the values of x into the equations one by one and calculate the corresponding values of y.
Let's start with the first equation: y = 2x^2 + 9x - 4.
For x = -2:
y = 2(-2)^2 + 9(-2) - 4
y = 8 - 18 - 4
y = -14
For x = -1:
y = 2(-1)^2 + 9(-1) - 4
y = 2 - 9 - 4
y = -11
For x = 0:
y = 2(0)^2 + 9(0) - 4
y = 0 + 0 - 4
y = -4
For x = 1:
y = 2(1)^2 + 9(1) - 4
y = 2 + 9 - 4
y = 7
For x = 2:
y = 2(2)^2 + 9(2) - 4
y = 8 + 18 - 4
y = 22
Now, let's move on to the second equation: y = -x^2 + 6x + 2.
For x = -2:
y = -(-2)^2 + 6(-2) + 2
y = -4 - 12 + 2
y = -14
For x = -1:
y = -(-1)^2 + 6(-1) + 2
y = -1 - 6 + 2
y = -5
For x = 0:
y = -(0)^2 + 6(0) + 2
y = 0 + 0 + 2
y = 2
For x = 1:
y = -(1)^2 + 6(1) + 2
y = -1 + 6 + 2
y = 7
For x = 2:
y = -(2)^2 + 6(2) + 2
y = -4 + 12 + 2
y = 10
Therefore, the values of y for the first equation y = 2x^2 + 9x - 4, and the second equation y = -x^2 + 6x + 2, within the range of x from -2 to 2 are:
(-2, -14), (-1, -11), (0, -4), (1, 7), and (2, 22) for the first equation,
and
(-2, -14), (-1, -5), (0, 2), (1, 7), and (2, 10) for the second equation.