Example 1.2 TTextbook p30, Example 21 Find the coordinates of the points of intersection between the curve \( y=x^{2}-2 x+2 \) and the line \( x+2 y=4 \).

To find the points of intersection between the curve \( y = x^{2} - 2x + 2 \) and the line \( x + 2y = 4 \), we'll solve the system of equations simultaneously. ### Step 1: Express One Variable in Terms of the Other From the linear equation: \[ x + 2y = 4 \] we can solve for \( x \) in terms of \( y \): \[ x = 4 - 2y \] ### Step 2: Substitute into the Quadratic Equation Substitute \( x = 4 - 2y \) into the quadratic equation \( y = x^{2} - 2x + 2 \): \[ y = (4 - 2y)^{2} - 2(4 - 2y) + 2 \] Expand and simplify: \[ y = (16 - 16y + 4y^{2}) - (8 - 4y) + 2 \] \[ y = 16 - 16y + 4y^{2} - 8 + 4y + 2 \] \[ y = 4y^{2} - 12y + 10 \] ### Step 3: Rearrange into a Standard Quadratic Form Bring all terms to one side to set the equation to zero: \[ 4y^{2} - 12y + 10 - y = 0 \] \[ 4y^{2} - 13y + 10 = 0 \] ### Step 4: Solve the Quadratic Equation Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = -13 \), and \( c = 10 \): \[ y = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 4 \cdot 10}}{2 \cdot 4} \] \[ y = \frac{13 \pm \sqrt{169 - 160}}{8} \] \[ y = \frac{13 \pm \sqrt{9}}{8} \] \[ y = \frac{13 \pm 3}{8} \] This gives two solutions: 1. \( y = \frac{16}{8} = 2 \) 2. \( y = \frac{10}{8} = \frac{5}{4} \) ### Step 5: Find the Corresponding \( x \) Values Using \( x = 4 - 2y \): 1. For \( y = 2 \): \[ x = 4 - 2(2) = 4 - 4 = 0 \] So, one point of intersection is \( (0, 2) \). 2. For \( y = \frac{5}{4} \): \[ x = 4 - 2\left(\frac{5}{4}\right) = 4 - \frac{10}{4} = 4 - \frac{5}{2} = \frac{8}{2} - \frac{5}{2} = \frac{3}{2} \] So, the other point of intersection is \( \left(\frac{3}{2}, \frac{5}{4}\right) \). ### **Final Answer** The curves intersect at the points \( (0,\ 2) \) and \( \left(\dfrac{3}{2},\ \dfrac{5}{4}\right) \).