What is \circ f(x)=(5x-2^2)(6x+1) ?

The solution to \circ f(x)=(5x-2^2)(6x+1) is Parabola with (h,k)=(19/60 ,-841/120),p= 1/120

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\circ f(x)=(5x-2^2)(6x+1)

\circ f (x) = (5 x - 2^{2}) (6 x + 1)

(h, k) = (\frac{19}{60}, - \frac{841}{120}), p = \frac{1}{120}

Solution steps

\circ y = (5 x - 2^{2}) (6 x + 1)

Rewrite

\circ y = (5 x - 2^{2}) (6 x + 1)

in the standard form

4 \cdot \frac{1}{120} (y - (- \frac{841}{120})) = (x - \frac{19}{60})^{2}

Therefore parabola properties are:

(h, k) = (\frac{19}{60}, - \frac{841}{120}), p = \frac{1}{120}

(y + 5)^{2} = - 5 (x - 9)

(y - 3)^{2} = 8 x

x^{2} + 4 x - 12 y - 56 = 0

x, x^{2} = 5 - 5 w

x^{2} \leq - 2 y

What is \circ f(x)=(5x-2^2)(6x+1) ?
The solution to \circ f(x)=(5x-2^2)(6x+1) is Parabola with (h,k)=(19/60 ,-841/120),p= 1/120