What is the vertices f(x)=-x^2-14x-48 ?

The vertices f(x)=-x^2-14x-48 is Maximum (-7,1)

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vertices f(x)=-x^2-14x-48

vertices

f (x) = - x^{2} - 14 x - 48

Maximum (- 7, 1)

Solution steps

y = - x^{2} - 14 x - 48

The parabola parameters are:

a = - 1, b = - 14, c = - 48

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{( - 14 )}{2 ( - 1 )}

Simplify

x_{v} = - 7

Plug in

x_{v} = - 7

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(- 7, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 7, 1)

x^{2} + 10 x - 2 y + 23 = 0

+ x^{2} - 36

f (x) = 2 (x + 14)^{2} - 6

y = 3 x^{2} - 6 x + 5

x^{2} - 2

What is the vertices f(x)=-x^2-14x-48 ?
The vertices f(x)=-x^2-14x-48 is Maximum (-7,1)