What is the vertices f(x)=-x^2+14x-48 ?

The vertices f(x)=-x^2+14x-48 is Maximum (7,1)

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vertices f(x)=-x^2+14x-48

vertices

f (x) = - x^{2} + 14 x - 48

Maximum (7, 1)

Solution steps

y = - x^{2} + 14 x - 48

The parabola parameters are:

a = - 1, b = 14, c = - 48

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

x_{v} = - \frac{14}{2 ( - 1 )}

Simplify

- \frac{14}{2 ( - 1 )} : 7

x_{v} = 7

Plug in

x_{v} = 7

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(7, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (7, 1)

f (x) = x^{2} + 11 x - 35

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

y = 3 x^{2} - 6 x + 4

y = 3 x^{2} + 2 x - 1

x^{2} - x

What is the vertices f(x)=-x^2+14x-48 ?
The vertices f(x)=-x^2+14x-48 is Maximum (7,1)