Intercepting curvesDemonstrate that the curves x^2-3x+1 and 2x^2+x+4 are intercepting.

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should know that the curves share a common point if the x coordinate of the point check both equations of curves, such that:

`x^2 - 3x + 1 = 2x^2 + x + 4`

`x^2 - 2x^2 - 3x - x + 1 - 4 = 0`

`-x^2 - 4x - 3 = 0 => x^2 + 4x + 3 = 0`

You should use factorization to solve quadratic equation, such that:

`x^2 + x + 3x + 3 = 0`

`x(x + 1) + 3(x + 1) = 0 => (x + 1)(x + 3) = 0 => {(x + 1 = 0),(x + 3 = 0):} => x = -1, x = -3`

Hence, the curves intersect each other at `x = -1` and `x = -3.`

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To prove that 2 curves are intercepting, we'll have to solve the system formed by the equations of the curves. If the curves are intercepting, then the system will have solutions.

We'll note y = x^2-3x+1 (1)

and y = 2x^2+x+4 (2)

We'll put (1) = (2)

x^2-3x+1 = 2x^2+x+4

We'll subtract boths sides x^2-3x+1 and we'll use symmetric property:

x^2 + 4x + 3 = 0

x1 = [-4 + sqrt(16 - 12)]/2

x1 = (-4+2)/2

x1 = -1

x2 = (-4-2)/2

x2 = -3

For x1 = -1, y1 = 1+3+1 = 5

For x2 = -3, y = 9+9+1 = 19

The intercepting points of the given curves are: (-1 ; 5) and (-3 ; 19).