What is the vertices f(x)=2x^2-3x-5 ?

The vertices f(x)=2x^2-3x-5 is Minimum (3/4 ,-49/8)

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vertices f(x)=2x^2-3x-5

vertices

f (x) = 2 x^{2} - 3 x - 5

Minimum (\frac{3}{4}, - \frac{49}{8})

Solution steps

y = 2 x^{2} - 3 x - 5

The parabola parameters are:

a = 2, b = - 3, c = - 5

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

x_{v} = - \frac{( - 3 )}{2 \cdot 2}

Simplify

x_{v} = \frac{3}{4}

Plug in

x_{v} = \frac{3}{4}

to find the

y_{v}

value

y_{v} = - \frac{49}{8}

Therefore the parabola vertex is

(\frac{3}{4}, - \frac{49}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (\frac{3}{4}, - \frac{49}{8})

y = - 2 x^{2} - 16 x - 37

x = 3 y^{2}

y = x^{2} + 4 x - 12

2 (x + 1)^{2} - 3

2 y^{2} - 8 x = 0

What is the vertices f(x)=2x^2-3x-5 ?
The vertices f(x)=2x^2-3x-5 is Minimum (3/4 ,-49/8)