2. error analysis cameron said that the y - intercept of a quadratic function always tells the maximum value…

Brief Explanations:

The ( y )-intercept of a quadratic function ( y = ax^{2}+bx + c) (found by setting ( x = 0), so ( y=c)) gives the point ((0,c)) where the graph crosses the ( y )-axis. The maximum (or minimum) value of a quadratic function occurs at the vertex. For a quadratic function (y = ax^{2}+bx + c), the (x) - coordinate of the vertex is (x=-\frac{b}{2a}), and the (y) - coordinate (which is the maximum if (a<0) or minimum if (a > 0)) is (y = a\left(-\frac{b}{2a}\right)^{2}+b\left(-\frac{b}{2a}\right)+c=\frac{4ac - b^{2}}{4a}). The (y) - intercept (c) has no direct relation to whether the function has a maximum (or minimum) value (which depends on the sign of (a)) or the value of the maximum (which depends on (a), (b), and (c) as shown in the vertex formula).

Answer:

Cameron's error is that the (y) - intercept ((c) in (y=ax^{2}+bx + c)) does not determine the maximum value of the quadratic function. The maximum value of a quadratic function (y = ax^{2}+bx + c) (when (a<0)) is at the vertex (\left(-\frac{b}{2a},\frac{4ac - b^{2}}{4a}\right)) and depends on (a), (b), and (c), not just (c) (the (y) - intercept). The (y) - intercept only gives the value of the function when (x = 0).