What is the vertices f(x)=-16x^2+4x+10 ?

The vertices f(x)=-16x^2+4x+10 is Maximum (1/8 , 41/4)

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vertices f(x)=-16x^2+4x+10

vertices

Maximum (\frac{1}{8}, \frac{41}{4})

Solution steps

y = - 16 x^{2} + 4 x + 10

The parabola parameters are:

a = - 16, b = 4, c = 10

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

x_{v} = - \frac{4}{2 ( - 16 )}

Simplify

- \frac{4}{2 ( - 16 )} : \frac{1}{8}

x_{v} = \frac{1}{8}

Plug in

x_{v} = \frac{1}{8}

to find the

y_{v}

value

y_{v} = \frac{41}{4}

Therefore the parabola vertex is

(\frac{1}{8}, \frac{41}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 16

Maximum (\frac{1}{8}, \frac{41}{4})

What is the vertices f(x)=-16x^2+4x+10 ?
The vertices f(x)=-16x^2+4x+10 is Maximum (1/8 , 41/4)