What is the vertices f(x)=x^2+11x-35 ?

The vertices f(x)=x^2+11x-35 is Minimum (-11/2 ,-261/4)

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vertices f(x)=x^2+11x-35

vertices

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

Minimum (- \frac{11}{2}, - \frac{261}{4})

Solution steps

y = x^{2} + 11 x - 35

The parabola parameters are:

a = 1, b = 11, c = - 35

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

x_{v} = - \frac{11}{2 \cdot 1}

Simplify

x_{v} = - \frac{11}{2}

Plug in

x_{v} = - \frac{11}{2}

to find the

y_{v}

value

y_{v} = - \frac{261}{4}

Therefore the parabola vertex is

(- \frac{11}{2}, - \frac{261}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{11}{2}, - \frac{261}{4})

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

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

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

x^{2} - x

y = - x^{2} + 6 x - 1

What is the vertices f(x)=x^2+11x-35 ?
The vertices f(x)=x^2+11x-35 is Minimum (-11/2 ,-261/4)