What is the vertices f(x)=-2(x-1)^2-5 ?

The vertices f(x)=-2(x-1)^2-5 is Maximum (1,-5)

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vertices f(x)=-2(x-1)^2-5

vertices

f (x) = - 2 (x - 1)^{2} - 5

Maximum (1, - 5)

Solution steps

y = - 2 (x - 1)^{2} - 5

Rewrite

y = - 2 (x - 1)^{2} - 5

in the form

y = a x^{2} + b x + c

y = - 2 x^{2} + 4 x - 7

The parabola parameters are:

a = - 2, b = 4, c = - 7

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

x_{v} = - \frac{4}{2 ( - 2 )}

Simplify

- \frac{4}{2 ( - 2 )} : 1

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 5

Therefore the parabola vertex is

(1, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (1, - 5)

f (x) = y^{2} + 3 y + 4

+ \frac{x ^{2} + x}{2}

f (x) = 4 x^{2} - 32 x - 7

2 (x + 1)^{2}

t^{2} + 12 t - 18

What is the vertices f(x)=-2(x-1)^2-5 ?
The vertices f(x)=-2(x-1)^2-5 is Maximum (1,-5)