Question

Hello, so for number#11 I used the vertex formula but got an incorrect answer for 'a'. I got a=-2, but I looked up the answer and its a=-1, how so?

Answer 1

Maximum (-1,5)

Explanation:

The vertex of an up - down facing parabola of the form

$y\text{ = }ax^2+bx+c\text{ }$

is

$x_v\text{ = -}\frac{b}{2a}$

. Quadratic form of the function

$y\text{ = }-2x^2-4x+3$

The parameters become

a = -2

b = -4

c = 3

$\begin{gathered} x_v\text{ = -}\frac{-4}{2(-2)} \\ x_v\text{ = -}\frac{-4}{-4} \\ x_v\text{ = -1} \end{gathered}$

To find y(v) simply put x(v) in the function

$\begin{gathered} y_v\text{ = -2*\lparen-1\rparen}^2\text{ - 4\lparen-1\rparen +3} \\ y_v\text{ = -2 + 4 +3} \\ y_v\text{ = 5} \end{gathered}$$\begin{gathered} Solution: \\ (x_v,y_v)\text{ = \lparen}-1,5) \end{gathered}$

NB:

If a < 0, then the vertex is a maximum value (Which is the case here since -2 < 0)

If a > 0, then the vertex is a minimum value