# Sketch the region enclosed by y= 6|x| and y = x^2 -7

• apiwowar
In summary: Completing the square, these are the same as (x- 3)^2- 16= 0 so (x- 3)^2= 16 or x- 3= 4 or x- 3= -4. That is, x= 7 or x= -1 so the solutions are a= -1, b= 7. In summary, to solve for x and find the limits of integration, set the two functions equal to each other and solve for x in two cases: x>=0 and x<0. The solutions are a= -1 and b= 7.
apiwowar
i know that the way to solve this is by evaluating the integral from a to b of the first function minus the second one but how would i solve for x to find out what the limits of integration should be?

if you set them equal to each other you get 6|x| = x^2 - 7
but I am not exactly sure what to do with the |x|

like rootX said, when dealing with absolute value functions, it's usually best to define it peace-wise.

f(x) = 6x for x >= 0
-6x for x < 0

Or you could notice that the two functions are even, and you could thus solve for x using 6x, and keep in mind that there is another intersection point opposite the y-axis.

apiwowar said:
i know that the way to solve this is by evaluating the integral from a to b of the first function minus the second one but how would i solve for x to find out what the limits of integration should be?

if you set them equal to each other you get 6|x| = x^2 - 7
but I am not exactly sure what to do with the |x|
If $x\ge 0$, |x|= x so this is 6x= x^2- 7 which is the same as x^2- 6x- 7= 0.

If x< 0, |x|= -x so this is -6x= x^2- 7 which is the same as x^2+ 6x- 7= 0.

## 1. What is the shape of the region enclosed by y= 6|x| and y = x^2 -7?

The region enclosed by these two equations is a combination of a parabola and two line segments. It resembles a "V" shape, with the vertex at the origin and the two arms extending upwards and downwards.

## 2. What are the x and y intercepts of this region?

The x-intercepts can be found by setting y=0 and solving for x. This gives us two points: (0,0) and (6,0). The y-intercept can be found by setting x=0, giving us a point at (0,-7).

## 3. How can I find the area of this region?

To find the area of this region, we need to split it into two parts: the area under the parabola and the area between the two line segments. We can use integration to find the area under the parabola, and then find the area of the two triangles formed by the line segments using the formula A = 1/2 * base * height. Finally, we add these two areas together to get the total area of the region.

## 4. What is the domain and range of this region?

The domain of this region is all real numbers, as both equations are defined for any value of x. The range, however, is limited. The lower bound of the range is -7, which is the y-intercept of the parabola. The upper bound is infinity, as the parabola and line segments extend infinitely upwards and downwards.

## 5. How does changing the coefficient of x in y= 6|x| affect the shape of the region?

Changing the coefficient of x in y= 6|x| affects the steepness of the two line segments that form the region. A larger coefficient will result in steeper lines, while a smaller coefficient will result in more gradual lines. The overall shape of the region will remain the same, but the size of the V will change.

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