What is the vertices y=x^2-3x^2 ?

The vertices y=x^2-3x^2 is Maximum (0,0)

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vertices y=x^2-3x^2

vertices

y = x^{2} - 3 x^{2}

Maximum (0, 0)

Solution steps

y = x^{2} - 3 x^{2}

Rewrite

y = x^{2} - 3 x^{2}

in the form

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

y = - 2 x^{2}

The parabola parameters are:

a = - 2, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( - 2 )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(0, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (0, 0)

f (x) = 8 x^{2} + 4 x - 1

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

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

+ x^{2} - 36

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