What is the vertices f(x)=4x^2-3 ?

The vertices f(x)=4x^2-3 is Minimum (0,-3)

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vertices f(x)=4x^2-3

vertices

f (x) = 4 x^{2} - 3

Minimum (0, - 3)

Solution steps

y = 4 x^{2} - 3

The parabola parameters are:

a = 4, b = 0, c = - 3

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

x_{v} = - \frac{0}{2 \cdot 4}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 3

Therefore the parabola vertex is

(0, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (0, - 3)

f (x) = - x^{2} + 4 x + 12

y^{2} = 12 x

y = \frac{x ^{2} x ^{1}}{x}

(x - 3)^{2} = 8 (y - 5)

y^{2} = - 24 x