What is the vertices+10x^2-15x+2 ?

The vertices+10x^2-15x+2 is Minimum (3/4 ,-29/8)

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vertices+10x^2-15x+2

vertices

+ 10 x^{2} - 15 x + 2

Minimum (\frac{3}{4}, - \frac{29}{8})

Solution steps

y = 10 x^{2} - 15 x + 2

The parabola parameters are:

a = 10, b = - 15, c = 2

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

x_{v} = - \frac{( - 15 )}{2 \cdot 10}

Simplify

- \frac{- 15}{2 \cdot 10} : \frac{3}{4}

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

Plug in

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

to find the

y_{v}

value

y_{v} = - \frac{29}{8}

Therefore the parabola vertex is

(\frac{3}{4}, - \frac{29}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 10

Minimum (\frac{3}{4}, - \frac{29}{8})

f (x) = 4 x^{2} + 8 x - 8

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

f (x) = 4 x^{2} + 8 x + 3

x = 7 y^{2}

y = - x^{2} + 10 x - 28

What is the vertices+10x^2-15x+2 ?
The vertices+10x^2-15x+2 is Minimum (3/4 ,-29/8)