What is the vertices f(x)=-9.5x^2-47.5x+63 ?

The vertices f(x)=-9.5x^2-47.5x+63 is Maximum (-2.5,122.375)

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vertices f(x)=-9.5x^2-47.5x+63

vertices

f (x) = - 9.5 x^{2} - 47.5 x + 63

Maximum (- 2.5, 122.375)

Solution steps

y = - 9.5 x^{2} - 47.5 x + 63

The parabola parameters are:

a = - 9.5, b = - 47.5, c = 63

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

x_{v} = - \frac{( - 47.5 )}{2 ( - 9.5 )}

Simplify

x_{v} = - 2.5

Plug in

x_{v} = - 2.5

to find the

y_{v}

value

y_{v} = 122.375

Therefore the parabola vertex is

(- 2.5, 122.375)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 9.5

Maximum (- 2.5, 122.375)

f (x) = 3 (x - 1) (x + 2)

f (x) = - \frac{1}{250} x^{2} + \frac{4}{5} x

y = 2 x^{2} - 8 x + 4

f (x) = 4 (x + 4)^{2} + 6

4 (x + 2)^{2} + 5

What is the vertices f(x)=-9.5x^2-47.5x+63 ?
The vertices f(x)=-9.5x^2-47.5x+63 is Maximum (-2.5,122.375)