What is the vertices y=7x^2-10x+5 ?

The vertices y=7x^2-10x+5 is Minimum (5/7 , 10/7)

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vertices y=7x^2-10x+5

vertices

y = 7 x^{2} - 10 x + 5

Minimum (\frac{5}{7}, \frac{10}{7})

Solution steps

y = 7 x^{2} - 10 x + 5

The parabola parameters are:

a = 7, b = - 10, c = 5

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

x_{v} = - \frac{( - 10 )}{2 \cdot 7}

Simplify

- \frac{- 10}{2 \cdot 7} : \frac{5}{7}

x_{v} = \frac{5}{7}

Plug in

x_{v} = \frac{5}{7}

to find the

y_{v}

value

y_{v} = \frac{10}{7}

Therefore the parabola vertex is

(\frac{5}{7}, \frac{10}{7})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 7

Minimum (\frac{5}{7}, \frac{10}{7})

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

(x + 2)^{2} - 9

y = x^{2} - 8 x + 7

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

y = - 4 x^{2} + 8 x + 2

What is the vertices y=7x^2-10x+5 ?
The vertices y=7x^2-10x+5 is Minimum (5/7 , 10/7)