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

The vertices x=y^2+3y+4 is (h,k)=(7/4 ,-3/2),p= 1/4

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vertices x=y^2+3y+4

vertices

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

(h, k) = (\frac{7}{4}, - \frac{3}{2}), p = \frac{1}{4}

Solution steps

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{4} (x - \frac{7}{4}) = (y - (- \frac{3}{2}))^{2}

Therefore parabola properties are:

(h, k) = (\frac{7}{4}, - \frac{3}{2}), p = \frac{1}{4}

y = 2 x^{2} - 5

y = 500 (0.04 - x^{2})

- 5 x^{2} - 40 x - 77

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

y = 5 x^{2} - 2 x - 1

What is the vertices x=y^2+3y+4 ?
The vertices x=y^2+3y+4 is (h,k)=(7/4 ,-3/2),p= 1/4