find the area enclosed by the line y= x -1 and parabola y^2=2x +6?

Question

my book says that by solving the two equations we find that the point of intersection are (-1, -2) and (5, 4). i tried solving the equations but have no idea how to get those numbers

Answer 1

Solve this:(x-1)^2=2x+6 i.e. x^2-4x-5=0, a=1,b=-4, c=-5
D=b^2-4ac=16+20=36
sqrt(D)=6
x1=(4+6)/2=5, x2=(4-6)/2=-1
y1=x1-1=5-1=4,y2=x2-1=-2.Then we have the points (-1,-2) and(5,4).the curve has equations :x=y+1 & x=1/2y^2-3 Then we have integral(y+1-1/2y^2+3)dy for y =-2 to4 = (4y+y^2/2-y^3/6) for y=-2 to 4=16+8-64/6-(-8+2+8/6)=30-56/6=124/6=62/3. END & GOOD LUCK!

Answer 2

You can find the points of intersection by plugging one equation into the other:

If y = x-1 and y^2 = 2x+6, then:

(x-1)^2 = 2x+6
x^2 -2x + 1 = 2x + 6
x^2 - 4x - 5 = 0
(x - 5)(x + 1) = 0

so the x values where the graph crosses are 5 and -1, and the corresponding y-values are y = 5-1 = 4 and y = -1-1 = -2. So the book's right, the intersection points are (-1, -2) and (5,4).

Your question doesn't ask about finding the area itself, so I guess I'll stop there.

Answer 3

Okay you have (x-1)^2=2x+6, or x^2-2x+1=2x+6, or x^2-4x-5.
So, factor it and you get (x+1)(x-5)=0.
So, x=-1, and x=5.

And just plug into y=x-1 to find y.

To find area, y=(2x+6)^.5.
SO, intgerate it, let u=2x+6., so du/dx=2, so dx=du/2.
So, integrate 1/2 S udu ,you get (1/2 * 2/3)u^3/2
So, you get (2x+6)^(3/2)/3.
Now area is (2*5+6)^(3/2)/3-(-1*5+6)^(3/2)/3 = 64/3 - 1/3 = 63/3=21.

Now integrate y=x-1, and you have x^2/2 - x.

So, area is 5^2/2 - 5 - (-1^2/2 - -1) = 7.5-1.5= = 6.
So, the area is 21-6=15.
(If I didnt make a mistake!)

Answer 4

sketch graphs on one coordinate gadget for seen help. (one is a horizontal parabola, the different a line) confirm factors of intersection (as those are the bounds as considered on the graph). you need to get (5,4) and (-a million,-2). write the two equations as purposes of y. it relatively is, sparkling up each and each for x. you need to have x = (y^2)/2 - 3 and x = y+a million combine with know to y. it relatively is, rather of making use of critical of [acceptable - backside], use critical of [suitable - left]. (you notice that x = y+a million is on the main suitable of the bounded area and x = (y^2)/2 - 3 is left. so... area = critical (from -2 to 4) of [y + a million - (y^2)/2 + 3] dy you need to wind up with 18 sq. gadgets

Answer 5

(x-1)^2 = 2x+6
or x^2 -2x + 1 = 2x + 6
or x^2 -4x -5 = 0
or (x-5)(x+1) = 0

There you go...........