What is -f(x)=4(x+1)^2+4(x+1)-3 ?

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

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-f(x)=4(x+1)^2+4(x+1)-3

- f (x) = 4 (x + 1)^{2} + 4 (x + 1) - 3

(h, k) = (- \frac{3}{2}, 4), p = - \frac{1}{16}

Solution steps

- y = 4 (x + 1)^{2} + 4 (x + 1) - 3

Rewrite

- y = 4 (x + 1)^{2} + 4 (x + 1) - 3

in the standard form

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

Therefore parabola properties are:

(h, k) = (- \frac{3}{2}, 4), p = - \frac{1}{16}

x^{2} + 4 y = 4

3 x = - 16 w (x, y) y

- 4 (x - 5) = (y - 4)^{2}

9 x^{2} - m (x) 2 = 0

2 - f (x) = x^{2} - 4 x - 5

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