Example 4 . \( \quad c \) A quadratic curve crosses the \( y \)-axis at \( (0,-10) \) and the \( x \)-axis at \( (1,0) \) and \( (k, 0) \) where \( k>1 \). The maximum point occurs at \( x=3 \). State the value of \( k \) and find the equation of the curve in the form \( y=a x^{2}+b x+c \).

To find the value of \( k \) and the equation of the quadratic curve, we can use the given information. 1. **Vertex Form**: The maximum point occurring at \( x = 3 \) suggests that the vertex form of the quadratic is \( y = a(x - 3)^2 + b \). Since it opens downwards (maximum point), \( a < 0 \). 2. **Intersection with the y-axis**: We know the curve crosses the y-axis at \( (0, -10) \). So we can plug in \( x = 0 \): \[ -10 = a(0 - 3)^2 + b \Rightarrow -10 = 9a + b \] 3. **Intersection with the x-axis**: The roots of the quadratic curve are given as \( (1,0) \) and \( (k, 0) \). Therefore, we can also use the fact that: \[ y = a(x - 1)(x - k) \] 4. **Using the vertex**: The x-coordinate of the vertex can also be calculated as: \[ \frac{1 + k}{2} = 3 \Rightarrow 1 + k = 6 \Rightarrow k = 5. \] 5. **Finding the equation**: Now we have \( k = 5 \). The roots are \( (1, 0) \) and \( (5, 0) \): \[ y = a(x - 1)(x - 5) = a(x^2 - 6x + 5). \] Now we need the value of \( a \). Since we know the curve crosses the y-axis at \( (0, -10) \): \[ -10 = a(0^2 - 6(0) + 5) \Rightarrow -10 = 5a \Rightarrow a = -2. \] 6. **Equation of the curve**: Plugging \( a \) back into the equation gives: \[ y = -2(x^2 - 6x + 5) = -2x^2 + 12x - 10. \] Thus, the solution is: - The value of \( k \) is \( 5 \). - The equation of the curve is \( y = -2x^2 + 12x - 10 \).