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Popular >

vertices 28x^2+47x+15

Solution

vertices

28 x^{2} + 47 x + 15

Solution

Minimum (- \frac{47}{56}, - \frac{529}{112})

Solution steps

y = 28 x^{2} + 47 x + 15

The parabola parameters are:

a = 28, b = 47, c = 15

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

x_{v} = - \frac{47}{2 \cdot 28}

Simplify

x_{v} = - \frac{47}{56}

Plug in

x_{v} = - \frac{47}{56}

to find the

y_{v}

value

y_{v} = - \frac{529}{112}

Therefore the parabola vertex is

(- \frac{47}{56}, - \frac{529}{112})

If

a < 0,

then the vertex is a maximum value

If

a > 0,

then the vertex is a minimum value

a = 28

Minimum (- \frac{47}{56}, - \frac{529}{112})

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Frequently Asked Questions (FAQ)

What is the vertices 28x^2+47x+15 ?
The vertices 28x^2+47x+15 is Minimum (-47/56 ,-529/112)