What is the vertices y=25x^2-36 ?

The vertices y=25x^2-36 is Minimum (0,-36)

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vertices y=25x^2-36

vertices

y = 25 x^{2} - 36

Minimum (0, - 36)

Solution steps

y = 25 x^{2} - 36

The parabola parameters are:

a = 25, b = 0, c = - 36

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

x_{v} = - \frac{0}{2 \cdot 25}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 36

Therefore the parabola vertex is

(0, - 36)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 25

Minimum (0, - 36)

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

y^{2} = 4 x

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

y = x^{2} + 2 x + 1

10 + 3 x^{2}