What is the vertices y=5x^2-3x-7+2x^2-x ?

The vertices y=5x^2-3x-7+2x^2-x is Minimum (2/7 ,-53/7)

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vertices y=5x^2-3x-7+2x^2-x

vertices

y = 5 x^{2} - 3 x - 7 + 2 x^{2} - x

Minimum (\frac{2}{7}, - \frac{53}{7})

Solution steps

y = 5 x^{2} - 3 x - 7 + 2 x^{2} - x

Rewrite

y = 5 x^{2} - 3 x - 7 + 2 x^{2} - x

in the form

y = a x^{2} + b x + c

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

The parabola parameters are:

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

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

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

Simplify

- \frac{- 4}{2 \cdot 7} : \frac{2}{7}

x_{v} = \frac{2}{7}

Plug in

x_{v} = \frac{2}{7}

to find the

y_{v}

value

y_{v} = - \frac{53}{7}

Therefore the parabola vertex is

(\frac{2}{7}, - \frac{53}{7})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 7

Minimum (\frac{2}{7}, - \frac{53}{7})

4 y^{2} - 48 x - 20 y = 71

y = x^{2} + 2 x - 15

f (x) = - 4 (x + 1)^{2} - 5

x^{2} - 4

2 x^{2} + 3 x - 20

What is the vertices y=5x^2-3x-7+2x^2-x ?
The vertices y=5x^2-3x-7+2x^2-x is Minimum (2/7 ,-53/7)