# Prove that the curves x^2-3x+1 and 2x^2+x+4 are intercepting.

### 4 Answers | Add Yours

To find if the two curves intercept or not, we equate them, as at the point of intersection the x and y coordinates are the same.

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

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

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

=> x^2 + 3x + x + 3 = 0

=> x ( x + 3) + 1 ( x + 3) = 0

=> (x +1 )(x +3) = 0

=> x = -1 and -3

At x = -1 , y = -1^2 + 3 +1 = 1 + 3 + 1 = 5

At x = -3 , y = 9 + 9 +1 = 19

**Therefore the points of intersection are ( -1 , 5) and ( -3, 19)**

Given the curves:

f(x) = x^2 -3x +1

g(x) = 2x^2 +x +4

We need to find the intersection points-if any.

If f(x) and g(x) are intercepting, then the intersection points will be when f(x) = g(x).

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

We will combine like terms.

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

Now we will factor.

==> (x+3)(x+1) = 0

==> x= -3 ==> f(-3)=f(-3) = 19

==> x= -1 ==> f(-1)=g(-1) = 5

Then, there are 2 intersection points.

**(-1, 5) and (-3,19).**

Let y = x^2-3x+1....(1) and y= 2x^2+x+4...(2) intersect at a point (x,y).

Then (x,y) is on both curves (1) and (2). So the ordinates are same.

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

=> 0 = -(x^2-3x+1)+ 2x^2+x+4

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

=> 0 = (x+3)(x+1).

So x+1 = 0, or x+3 = 0.

=> x= -1, or x= -3.

So when x= -1, y =x^2-3x+1 = 1+3+1 = 5

When x= -3, y = x^2-3x+1 = 9+9+1 = 19

Therefore the curves are intersecting at (x,y) = (-1, 5) and (-3, 19).

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

We'll apply quadratic formula:

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).**