+109 What are the coordinates of the points where the graphs of f(x)=x^3 + x^2 - 3x + 5 and g(x) = x^3 + 2x^2 intersect?
Give your answer as a list of points separated by commas, with the points ordered such that their -coordinates are in increasing order. (So "(1,-3), (2,3), (5,-7)" - without the quotes - is a valid answer format.)
+98 Alright! I geuss I can try to solve this question!
First, let's note that the only points where the lines intersect is when we have \(f(x) = g(x)\).
This means that we only have to solve the equation \(x^3 + x^2 -3x + 5 = x^3 + 2x^2 \) to find the points where the lines intersect.
Combining like terms and moving all the terms to one side, we get the equation \(x^2 + 3x - 5 = 0\).
We can't factor this polynomial with integers, so the next best thing to do is to complete the square and find x that way.
We set up the equation by moving 5 to the other side, getting us \(x^2 + 3x = 5\). Now we complete the square by adding 9/4 to both sides.
\(x^2 + 3x + 9/4 = 5 + 9/4 \)
\((x + 3/2)^2 = 29/4\)
Now we just have to square root both sides to get us \(x+3/2= \pm \sqrt{29/4}\)
This means that \(x = \frac{\sqrt{29}-3}{2}, \frac{-\sqrt{29}-3}{2}\).
Subsituting these x values back into the g(x) and f(x) functions, we get \(y = \frac{-\sqrt{29}-3} 2, \frac{\sqrt{29}-3}{2}\)
So for the final answer, we have \((\frac{\sqrt{29}-3}{2}, \frac{-\sqrt{29}-3}{2}), (-\frac{\sqrt{29}-3}{2}, \frac{\sqrt{29}-3}{2})\)
Thanks! :) :) :)
