\( y=-2 x^{2}+10 x-8 \) and \( y=x^{2}-2 x-8 \)

Sure, let's analyze the two quadratic equations you've provided: 1. \( y = -2x^{2} + 10x - 8 \) 2. \( y = x^{2} - 2x - 8 \) Assuming you want to find the points where these two curves intersect, we'll set the equations equal to each other and solve for \( x \). ### Step 1: Set the Equations Equal to Each Other \[ -2x^{2} + 10x - 8 = x^{2} - 2x - 8 \] ### Step 2: Move All Terms to One Side to Form a Quadratic Equation Subtract \( x^{2} - 2x - 8 \) from both sides: \[ -2x^{2} + 10x - 8 - (x^{2} - 2x - 8) = 0 \] \[ -2x^{2} + 10x - 8 - x^{2} + 2x + 8 = 0 \] \[ -3x^{2} + 12x = 0 \] ### Step 3: Simplify the Equation \[ -3x^{2} + 12x = 0 \] Factor out the common term: \[ -3x(x - 4) = 0 \] ### Step 4: Solve for \( x \) Set each factor equal to zero: 1. \( -3x = 0 \) ⟹ \( x = 0 \) 2. \( x - 4 = 0 \) ⟹ \( x = 4 \) ### Step 5: Find the Corresponding \( y \)-Values Plug each \( x \)-value back into either of the original equations to find \( y \). **For \( x = 0 \):** Using \( y = -2x^{2} + 10x - 8 \): \[ y = -2(0)^{2} + 10(0) - 8 = -8 \] So, one intersection point is \( (0, -8) \). **For \( x = 4 \):** Using \( y = -2x^{2} + 10x - 8 \): \[ y = -2(4)^{2} + 10(4) - 8 = -2(16) + 40 - 8 = -32 + 40 - 8 = 0 \] So, the other intersection point is \( (4, 0) \). ### Conclusion The two quadratic equations intersect at the following points: - **Point 1:** \( (0, -8) \) - **Point 2:** \( (4, 0) \) If you need further analysis, such as graphing these equations or exploring their properties, feel free to ask!