# Determine the intercepting points between the line f(x) = x + 4 and the parabola f(x)= x^2 + 3x + 1

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f(x) = x+ 4

g(x) = x^2 + 3x + 1

We can find the intersection point when f(x) = g(x)

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

Now group similar:

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

Now factor:

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

==> x1= -3

==> x2= 1

Then y1= f(x1) = f(-3) = -3 + 4 = 1

and y2= f(x2)= f(1)= 1+ 4 = 5

Then we have two intercepting points:

( -3, 1) and (1, 5)

Then f and g intersects at two points:

(-3, 1) and (1,5)

In order to determine the intercepting points between the graphs of the given functions, we'll have to solve the system formed by the equations of the functions.

y = x^2 + 3x + 1

y = x + 4

We'll solve the system using substitution method.

We'll substitute y, into the second equation, by it's expression from the first equation.

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

We'll subtract x+4 both sides:

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

x^2 + 2x - 3 = 0

We'll apply the quadratic formula:

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

x1 = (-2+4)/2

**x1 = 1**

x2 = (-2-4)/2

**x2 = -3**

Now, we'll find y1 and y2.

y1 = x1 + 4

y1 = 1+4

**y1 = 5**

y2 = x2 + 4

y2 = -3+4

**y2 = 1**

The intercepting points are the solutions of the system: A(1,5); B (-3,1).

To find the interceping points of f(x) =x+4 and f(x) = x^2+3x+1.

Solution:

y = x+4 and y = x^2+3x+4. The y coordinates at the incepting points is same. So,

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

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

x^2+2x-3 = 0

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

x =1 or x=-3.

So from the 1st equation y = x+4 = 1+4 =5 when x=1.

y = x+4=-3+4 = 1 when x=-3.

So (1,5) and (-3, 1) are the intercepting points of f(x) = x+4 and f(x) = x^2+3x+1.