What is the vertices y=4x^2+16x+36 ?

The vertices y=4x^2+16x+36 is Minimum (-2,20)

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vertices y=4x^2+16x+36

vertices

y = 4 x^{2} + 16 x + 36

Minimum (- 2, 20)

Solution steps

y = 4 x^{2} + 16 x + 36

The parabola parameters are:

a = 4, b = 16, c = 36

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

x_{v} = - \frac{16}{2 \cdot 4}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 20

Therefore the parabola vertex is

(- 2, 20)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (- 2, 20)

y = \frac{1}{3} x - \frac{13}{3}

r (x) = (x - 8)^{2}

\frac{1}{2} x + 3

f (x) = 11 - 2 x

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

What is the vertices y=4x^2+16x+36 ?
The vertices y=4x^2+16x+36 is Minimum (-2,20)