What is 4x^2+7x-k(x)=0 ?

The solution to 4x^2+7x-k(x)=0 is Parabola with (h,k)=(-7/8 ,-49/16),p= 1/16

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4x^2+7x-k(x)=0

4 x^{2} + 7 x - k (x) = 0

(h, k) = (- \frac{7}{8}, - \frac{49}{16}), p = \frac{1}{16}

Solution steps

4 x^{2} + 7 x - y = 0

Rewrite

4 x^{2} + 7 x - y = 0

in the standard form

4 \cdot \frac{1}{16} (y - (- \frac{49}{16})) = (x - (- \frac{7}{8}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{7}{8}, - \frac{49}{16}), p = \frac{1}{16}

x - z^{2} = 0

x^{2} - 2 x - k (x) = 0

\frac{( x - 6 ) ^{2}}{16} + \frac{( y + 3 )}{25} = 1

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

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

What is 4x^2+7x-k(x)=0 ?
The solution to 4x^2+7x-k(x)=0 is Parabola with (h,k)=(-7/8 ,-49/16),p= 1/16