To find the points of intersection for the functions \( y = x^2 - 3x + 2 \) and \( y = 2x - 5 \), we need to set the equations equal to each other:
\[ x^2 - 3x + 2 = 2x - 5 \]
Rearranging the equation gives us:
\[ x^2 - 3x - 2x + 2 + 5 = 0 \]
This simplifies to:
\[ x^2 - 5x + 7 = 0 \]
Next, we can apply the quadratic formula to find the values of \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -5 \), and \( c = 7 \):
\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(7)}}{2(1)} \] \[ x = \frac{5 \pm \sqrt{25 - 28}}{2} \] \[ x = \frac{5 \pm \sqrt{-3}}{2} \]
Since the discriminant \( 25 - 28 = -3 \) is negative, there are no real solutions for \( x \). This means that the functions \( y = x^2 - 3x + 2 \) and \( y = 2x - 5 \) do not intersect at any real points; their graphs do not cross.
Thus, there are no intersection points, and the coordinates of the intersection points are non-existent (or can be said to have no real solutions).