# Intercepting point.What is the intercepting point of y = 2x^2 - 7 and the line y= -2x + 1?

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### 2 Answers

At the intercepting point of y = 2x^2 - 7 and the line y= -2x + 1 the values of x and y are the same.

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

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

=> x^2 + x - 4 = 0

Solving, x1 = -1/2 + sqrt (1 + 16) / 2

=> -1/2 + (sqrt 17)/2

x2 = -1/2 - (sqrt 17)/2

y1 = -2*x + 1 = 2 - (sqrt 17)

y2 = 2 + (sqrt 17)

**The points of intersection are(-1/2 + (sqrt 17)/2, 2 - sqrt 17) and (-1/2 - (sqrt 17)/2, 2 + sqrt 17)**

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).