What is the vertices f(x)f(x)=x^2+5x+6 ?

The vertices f(x)f(x)=x^2+5x+6 is (-2,0),(-3,0)

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vertices f(x)f(x)=x^2+5x+6

vertices

f (x) f (x) = x^{2} + 5 x + 6

(- 2, 0), (- 3, 0)

Solution steps

(h + a, k), (h - a, k)

Calculate Hyperbola properties

yy = x^{2} + 5 x + 6 :

Right-left Hyperbola with

(h, k) = (- \frac{5}{2}, 0), a = \frac{1}{2}, b = \frac{1}{2}

(- \frac{5}{2} + \frac{1}{2}, 0), (- \frac{5}{2} - \frac{1}{2}, 0)

Refine

(- 2, 0), (- 3, 0)

\frac{( x + 4 ) ^{2}}{9} - \frac{( y - 8 ) ^{2}}{49} = 1

25 x^{2} - 9 y^{2} - 225 = 0

5 x^{2} - 4 y^{2} - 40 x + 24 y + 24 = 0

x^{2} - y^{2} - 4 x - 6 y - 14 = 0

x^{2} - 16 y^{2} = 64

What is the vertices f(x)f(x)=x^2+5x+6 ?
The vertices f(x)f(x)=x^2+5x+6 is (-2,0),(-3,0)