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Where does the curve defined by f(x) = x^2 + 2x-1 and the line y= 2x+3 intersect?

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clara2 | Student, Undergraduate | eNoter

Posted January 5, 2011 at 11:19 AM via web

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Where does the curve defined by f(x) = x^2 + 2x-1 and the line y= 2x+3 intersect?

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted January 5, 2011 at 11:22 AM (Answer #1)

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Given the function f(x) = x^2 + 2x -1

and the line y= 2x+3

We need to find the intersection points for the curve and the line.

Then, we need to find the point that verifies f(x) and y.

==> f(x) = y

==> x^2 + 2x -1 = 2x +3

We will subtract 2x from both sides:

==> x^2 -1 = 3

Now we will add 1 to both sides.

==> x^2 = 4

==> x = +-2

Then, there are two points of intersection between the curve f(x) and the line y.

==> f(2) = 4+4-1 = 7

==> f(-2) = 4-4-1 = -1

Then, the points of intersection are:

(-2, -1) and (2,7)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted January 5, 2011 at 11:25 AM (Answer #2)

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To find the point of intersection of f(x) = x^2 + 2x-1 and the line y= 2x+3, we equate the two.

x^2 + 2x - 1 = 2x + 3

=> x^2 - 4 = 0

=> (x - 2)(x +2) = 0

x is equal to 2 and -2.

The corresponding values for y are 7 and -1

Therefore the points of intersection are (2, 7) and ( -2, -1)

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neela | High School Teacher | Valedictorian

Posted January 5, 2011 at 11:32 AM (Answer #3)

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To find the intersection point of y = x^2+x+1 (1) and y = 2x+3...(2).

We substitute f(x) = x^2+2x-1 in  place of y in (2):

 x^2+2x-1 = 2x+3.

We subtract 2x+3 from both sides:

x^2+2x-1 - 2x-3 = 0.

 x^2 -4 = 0.

 x^2 = 4.

We take the square root.

x = 2, or x= -2.

When x= 2, we get y = 2x+3, or  y = 2*2+3 = 7.

When x= -2, we get y = 2x+3 = 2*-2+3 = -1.

So the point of intersection of the curves are (2, 7) and (-2, -1). 

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