What is 4x^2+3x+h(x)=0 ?

The solution to 4x^2+3x+h(x)=0 is Parabola with (h,k)=(-3/8 , 9/16),p=-1/16

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4x^2+3x+h(x)=0

4 x^{2} + 3 x + h (x) = 0

(h, k) = (- \frac{3}{8}, \frac{9}{16}), p = - \frac{1}{16}

Solution steps

4 x^{2} + 3 x + y = 0

Rewrite

4 x^{2} + 3 x + y = 0

in the standard form

4 (- \frac{1}{16}) (y - \frac{9}{16}) = (x - (- \frac{3}{8}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{8}, \frac{9}{16}), p = - \frac{1}{16}

2 x^{2} + 3 x + k (x) = 0

\frac{x}{25} - y^{2} = 1

x = \frac{y ^{2}}{30} - 1

y^{2} = c^{1} (\frac{x + 1}{4 c ^{1}})

y \leq - \frac{1}{4} (x + 6)^{2} + 2

What is 4x^2+3x+h(x)=0 ?
The solution to 4x^2+3x+h(x)=0 is Parabola with (h,k)=(-3/8 , 9/16),p=-1/16