c) Draw the function \( f(x)=1-3 x+x^{2} \) where \( x \in[-1,4] \), then find: \( \begin{array}{ll}\text { i) the vertex of the curve } & \text { ii) max. point of } f(x) \\ \text { iii) equation of axis of symmetry. }\end{array} \)

To draw the function \( f(x) = 1 - 3x + x^2 \) within the interval \( x \in [-1, 4] \), first, let's recognize that it’s a quadratic function represented in the standard form \( ax^2 + bx + c \). The vertex of a parabola is a critical point that can be found using the formula \( x = -\frac{b}{2a} \). For this function, \( a = 1 \) and \( b = -3 \), so the x-coordinate of the vertex is \( x = -\frac{-3}{2 \cdot 1} = \frac{3}{2} = 1.5 \). To find the vertex, substitute \( x = 1.5 \) into the function to find the y-value: \( f(1.5) = 1 - 3(1.5) + (1.5)^2 = 1 - 4.5 + 2.25 = -1.25 \). Thus, the vertex is \( (1.5, -1.25) \). Now, regarding the maximum point of \( f(x) \), since the parabola opens upwards (positive \( a \)), the vertex is the minimum point, not a maximum. The minimum occurs at \( (1.5, -1.25) \), and the function continues to increase as \( x \) moves away from 1.5. Thus there is no maximum point in this interval. For the equation of the axis of symmetry, it coincides with the x-coordinate of the vertex. Therefore, the axis of symmetry is \( x = 1.5 \). In summary: i) Vertex: \( (1.5, -1.25) \) ii) Maximum point: None (the minimum point is at the vertex) iii) Equation of axis of symmetry: \( x = 1.5 \)