The graph of a quadratic model with a negative leading coefficient will have a maximum value at its vertex?

Question

explain

Answer 1

This will be a parabola in the shape of an arch, with general formula

y = - a^2 (x - b)^2 + c.

Here the vertex is at x = b; the maximum value that y achieves is c.

Live long and prosper.

Answer 2

The general form of a quadratic equation is
y = ax^2 + bx + c
Converting to vertex form,
y = a(x^2 + (b/a)x + c/a)
y = a(x^2 + (b/a)x + (b/(2a))^2 - (b/(2a))^2) + c
y = a(x + b/(2a))^2 - ab^2/(4a^2) + c
y = a(x + b/(2a))^2 + (4ac- b^2)/(4a)
Regardless of the signs of a and b, (x + b/(2a))^2 is minimum (0) when x = - b/(2a), so, when a is negative you are subtracting nothing from the quantity (4ac- b^2)/(4a) (which is the vertex), making this value the maximum of the function.

Answer 3

True. It will be in the shape of an upside down U. An example is y = -(x^2).