What is the vertices y=3x^2-5x7 ?

The vertices y=3x^2-5x7 is Minimum (35/6 ,-1225/12)

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vertices y=3x^2-5x7

vertices

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

Minimum (\frac{35}{6}, - \frac{1225}{12})

Solution steps

y = 3 x^{2} - 5 x \cdot 7

Rewrite

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

in the form

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

y = 3 x^{2} - 35 x

The parabola parameters are:

a = 3, b = - 35, c = 0

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

x_{v} = - \frac{( - 35 )}{2 \cdot 3}

Simplify

x_{v} = \frac{35}{6}

Plug in

x_{v} = \frac{35}{6}

to find the

y_{v}

value

y_{v} = - \frac{1225}{12}

Therefore the parabola vertex is

(\frac{35}{6}, - \frac{1225}{12})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (\frac{35}{6}, - \frac{1225}{12})

What is the vertices y=3x^2-5x7 ?
The vertices y=3x^2-5x7 is Minimum (35/6 ,-1225/12)