What is the vertices f(x)=y^2+3y+4 ?

The vertices f(x)=y^2+3y+4 is (h,k)=(7/(4f),-3/2),p= f/4

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vertices f(x)=y^2+3y+4

vertices

f (x) = y^{2} + 3 y + 4

(h, k) = (\frac{7}{4 f}, - \frac{3}{2}), p = \frac{f}{4}

Solution steps

f (x) = y^{2} + 3 y + 4

Rewrite

f x = y^{2} + 3 y + 4

in the standard form

4 \cdot \frac{f}{4} (x - \frac{7}{4 f}) = (y - (- \frac{3}{2}))^{2}

Therefore parabola properties are:

(h, k) = (\frac{7}{4 f}, - \frac{3}{2}), p = \frac{f}{4}

+ \frac{x ^{2} + x}{2}

f (x) = 4 x^{2} - 32 x - 7

2 (x + 1)^{2}

t^{2} + 12 t - 18

y = - 2 x^{2} - 4 x - 1

What is the vertices f(x)=y^2+3y+4 ?
The vertices f(x)=y^2+3y+4 is (h,k)=(7/(4f),-3/2),p= f/4