What is the vertices f(x)=2x^2+28x+105 ?

The vertices f(x)=2x^2+28x+105 is Minimum (-7,7)

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vertices f(x)=2x^2+28x+105

vertices

f (x) = 2 x^{2} + 28 x + 105

Minimum (- 7, 7)

Solution steps

y = 2 x^{2} + 28 x + 105

The parabola parameters are:

a = 2, b = 28, c = 105

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

x_{v} = - \frac{28}{2 \cdot 2}

Simplify

x_{v} = - 7

Plug in

x_{v} = - 7

to find the

y_{v}

value

y_{v} = 7

Therefore the parabola vertex is

(- 7, 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 7, 7)

4 x^{2} - 9 y^{2} - 36 = 0

x^{2} = 10 y

y^{2} + 2 y + x = 0

\frac{x ^{2}}{9} - \frac{y ^{2}}{25} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1

What is the vertices f(x)=2x^2+28x+105 ?
The vertices f(x)=2x^2+28x+105 is Minimum (-7,7)