# What is the intersepting point of y = 2x^2 - 7 and the line y= -2x + 1

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### 3 Answers

y= 2x^2 - 7

y= -2x + 1

the intersection point of both functions is where both kines have the same a and y coordinates:

==> y 1= y2

==> 2x^2 - 7 = -2x + 1

Now we will move all terms to the left side:

==> 2x^2 - 7 + 2x - 1 = 0

==> 2x^2 + 2x - 8 = 0

Now divide by 2:

==? x^2 + x - 4 = 0

==> x1= ( -1 + sqrt( 1+ 16) / 2 = ( -1 + sqrt17)/2

==? x2= ( 1- sqrt17) / 2

Now we will substitue in y2 to find y values:

y1= - 2x + 1 =-2 ( -1+ sqrt17)/2 + 1 = -sqrt17 + 2

y2 -2( -1-sqrt17)/2 + 1 = sqrt17 + 2

Then the points of intersections are:

(** -1+sqrt17)/2 , 2-sqrt17) and (( -1-sqrt17)/2 , sqrt17+2)**

To determine the intercepting point of the line and parabola, we'll have to solve the system:

2x^2 - y = 7 (1)

2x + y = 1 (2)

We'll calculate x form the 2nd equation:

2x = 1 - y

x = (1-y)/2

We'll substitute x in (1):

2*(1-y)^2/4 - y = 7

(1-y)^2/2 - y = 7

We'll multiply by 2 both sides:

(1-y)^2 - 2y - 14 = 0

We'll expand the square:

1 - 2y + y^2 - 2y - 14 = 0

We'll combine like terms and we'll re-arrange the terms:

y^2 - 4y - 13 = 0

We'll apply the quadratic:

y1=[4+sqrt(16+52)]/2

y1 = 2 + sqrt17

y2 = 2 - sqrt17

x1 = (1-y1)/2

x1 = (-1-sqrt17)/2

x2 = (-1+sqrt17)/2

**The intercepting points are:**

**((-1-sqrt17)/2 , 2 + sqrt17) and ((-1+sqrt17)/2 , 2 - sqrt17).**

To find the intersepting point of y = 2x^2 - 7 and the line y= -2x + 1:

The intersecting point is on both curves. So the ordinate y at the intersection point is same:

y =x^2-7 and y = -2x+1.

Therefore we equate the right side:

2x^2-7 = -2x+1.

2x^2+2x-7-1 = 0.

2x^2+2x-8 = 0.

We dvide by 2:

x^2+x-2 = 0.

(x+2)(x-1) = 0.

x = -2 or x= 1 .

So substitute the values x= -2 and x = 1 in y = 2x^2-7:

y = 2(-2)^2-7 = 1 and y = 2(1)^2 -7 = -5.

Therefore (-2,1) and (1,-5) are intersecting points of the curves.