Answer:
(0.618,4.236) and (-1.618,-0.236)
Step-by-step explanation:
To find the intersection, we are looking for a common point between the curves.
We are solving the system:
[tex]y=x^2+3x+2[/tex]
[tex]y=2x+3[/tex].
I'm going to do this by substitution:
[tex]x^2+3x+2=2x+3[/tex]
Subtract 2x and 3 on both sides:
[tex]x^2+1x-1=0[/tex]
[tex]x^2+x-1=0[/tex]
To solve this equation I'm going to use the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
To find [tex]a,b,\text{ and }c[/tex], you must compare [tex]x^2+x-1=0[/tex]
to [tex]ax^2+bx+c=0[/tex].
[tex]a=1,b=1,c=-1[/tex].
Now inputting the values into the quadratic formula gives us:
[tex]x=\frac{-1\pm\sqrt{(1)^2-4(1)(-1)}}{2(1)}[/tex]
[tex]x=\frac{-1\pm\sqrt{1+4}}{2}[/tex]
[tex]x=\frac{-1\pm\sqrt{5}}{2}[/tex]
This means you have two solutions:
[tex]x=\frac{-1+\sqrt{5}}{2} \text{ or } x=\frac{-1-\sqrt{5}}{2}[/tex]
It does say approximately.
So I'm going to put both of these in my calculator and I guess round to the nearest thousandths.
[tex]x=0.618 \text{ or } x=-1.618[/tex]
Now to find the corresponding y coordinates, I need to use one the equations along with each x.
I choose the linear equation: y=2x+3.
y=2x+3 when x=0.618
y=2(0.618)+3=4.236
So one approximate point is (0.618,4.236).
y=2x+3 when x=-1.618
y=2(-1.618)+3=-0.236
So another approximate point is (-1.618,-0.236).