What is the vertices f(x)=x^2+7x+12 ?

The vertices f(x)=x^2+7x+12 is Minimum (-7/2 ,-1/4)

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vertices f(x)=x^2+7x+12

vertices

f (x) = x^{2} + 7 x + 12

Minimum (- \frac{7}{2}, - \frac{1}{4})

Solution steps

y = x^{2} + 7 x + 12

The parabola parameters are:

a = 1, b = 7, c = 12

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

x_{v} = - \frac{7}{2 \cdot 1}

Simplify

x_{v} = - \frac{7}{2}

Plug in

x_{v} = - \frac{7}{2}

to find the

y_{v}

value

y_{v} = - \frac{1}{4}

Therefore the parabola vertex is

(- \frac{7}{2}, - \frac{1}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{7}{2}, - \frac{1}{4})

What is the vertices f(x)=x^2+7x+12 ?
The vertices f(x)=x^2+7x+12 is Minimum (-7/2 ,-1/4)