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

The vertices 5x^2+7y=3 is Maximum (0, 3/7)

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vertices 5x^2+7y=3

vertices

5 x^{2} + 7 y = 3

Maximum (0, \frac{3}{7})

Solution steps

5 x^{2} + 7 y = 3

Rewrite

5 x^{2} + 7 y = 3

in the form

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

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

The parabola parameters are:

a = - \frac{5}{7}, b = 0, c = \frac{3}{7}

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

x_{v} = - \frac{0}{2 ( - \frac{5}{7} )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = \frac{3}{7}

Therefore the parabola vertex is

(0, \frac{3}{7})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{5}{7}

Maximum (0, \frac{3}{7})

y = x^{2} + 2 x - 12

f (x) = 2 x^{2} - 80 x + 900

y = 5 x^{2}

f (x) = - (x + 3)^{2} + 5

4 t^{2}