# If f(x) = x^2 - x - 3. g(x) = 2x+3, what are the intersection points for f(x) and g(x).

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### 2 Answers

We have f(x) = x^2 - x - 3, g(x) = 2x+3.

At the point of intersection

f(x) = g(x)

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

=> x^2 - x - 2x - 3 - 3 = 0

=> x^2 - 3x - 6 = 0

Now x1 = [-b + sqrt (b^2 - 4ac)]/2a

=> [ 3 + sqrt ( 9 + 24)]/2

=> [3/2 + sqrt 33/2]

x2 = 3/2 - sqrt 33/2)

For x1, g(x) = 2*[3/2 + sqrt 33/2] +3

=> 6 + sqrt 33

For x2, g(x) = 2*[3/2 - sqrt 33/2] +3

=> 6 - sqrt 33

**So the points of contact are (([3/2 + sqrt 33/2]), (6 + sqrt 33)) and (([3/2 - sqrt 33/2]), (6 - sqrt 33))**

Given the foundations:

f(x) = x^2 - x - 3.

g(x) = 2x + 3

We need to find the intersection points of f and g.

We know that the intersection points are the values of x such that f(x) = g(x).

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

Now we will combine like terms.

==> x^2 -x -2x -3 -3 = 0

==> x^2 - 3x - 6 = 0

==> x1= (3 + sqrt(9+24) / 2 = ( 3+ sqrt(33) / 2

==> x1= ( 3+sqrt33)/2

==> x2= ( 3-sqrt33)/2

Then there are two intersection points for f and g.

==> g(3+sqrt33)/2) = 2(3+sqrt33)/2 + 3 = 6+sqrt33

==> g(3-sqrt33)/2) = 2(3-sqrt33)/2 + 3 = 6-sqrt33

Then, the points of intersection are:

**( (3+sqrt33)/2 , 6+sqrt33) and ( (3-sqrt33)/2 , 6-sqrt33)**