What is the vertices y=5x^2+13x-6 ?

The vertices y=5x^2+13x-6 is Minimum (-13/10 ,-289/20)

Popular >

vertices y=5x^2+13x-6

vertices

y = 5 x^{2} + 13 x - 6

Minimum (- \frac{13}{10}, - \frac{289}{20})

Solution steps

y = 5 x^{2} + 13 x - 6

The parabola parameters are:

a = 5, b = 13, c = - 6

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

x_{v} = - \frac{13}{2 \cdot 5}

Simplify

x_{v} = - \frac{13}{10}

Plug in

x_{v} = - \frac{13}{10}

to find the

y_{v}

value

y_{v} = - \frac{289}{20}

Therefore the parabola vertex is

(- \frac{13}{10}, - \frac{289}{20})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 5

Minimum (- \frac{13}{10}, - \frac{289}{20})

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

x^{2} - 12 x + 44

2 y^{2} - 8 x - 24 y + 56 = 0

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

f (x) = x^{2} + 3 x + \frac{1}{4}

What is the vertices y=5x^2+13x-6 ?
The vertices y=5x^2+13x-6 is Minimum (-13/10 ,-289/20)