What is the vertices f(x)=4x^2-7x-2 ?

The vertices f(x)=4x^2-7x-2 is Minimum (7/8 ,-81/16)

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vertices f(x)=4x^2-7x-2

vertices

f (x) = 4 x^{2} - 7 x - 2

Minimum (\frac{7}{8}, - \frac{81}{16})

Solution steps

y = 4 x^{2} - 7 x - 2

The parabola parameters are:

a = 4, b = - 7, c = - 2

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

x_{v} = - \frac{( - 7 )}{2 \cdot 4}

Simplify

x_{v} = \frac{7}{8}

Plug in

x_{v} = \frac{7}{8}

to find the

y_{v}

value

y_{v} = - \frac{81}{16}

Therefore the parabola vertex is

(\frac{7}{8}, - \frac{81}{16})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (\frac{7}{8}, - \frac{81}{16})

2 y^{2} + 8 y + x + 1 = 0

y^{2} - 12 y - 4 x + 28 = 0

f (x) = (x - 2)^{2} - 9

(3 x - 4) (4 x - 3) - 2 x

4 x^{2}

What is the vertices f(x)=4x^2-7x-2 ?
The vertices f(x)=4x^2-7x-2 is Minimum (7/8 ,-81/16)