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

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

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4x^2+x+m(x)=0

4 x^{2} + x + m (x) = 0

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

Solution steps

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

Rewrite

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

in the standard form

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

Therefore parabola properties are:

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

(x + 5)^{2} = 8 (y + 7)

3 X^{2} - 9 X - 2 = 5 Y

x^{2} = - \frac{20}{6} y

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

x^{2} + (y + 5) 2 = 99

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