What is x^2-5x+4-M>= 0 ?

The solution to x^2-5x+4-M>= 0 is Parabola with (h,k)=(5/2 ,-9/4),p= 1/4

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x^2-5x+4-M>= 0

x^{2} - 5 x + 4 - M \geq 0

(h, k) = (\frac{5}{2}, - \frac{9}{4}), p = \frac{1}{4}

Solution steps

x^{2} - 5 x + 4 - M \geq 0

Rewrite

x^{2} - 5 x + 4 - M \geq 0

in the standard form

4 \cdot \frac{1}{4} (M - (- \frac{9}{4})) = (x - \frac{5}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{5}{2}, - \frac{9}{4}), p = \frac{1}{4}

x = y - 5.3 x^{2}

y = - (x - 6)^{2} + 16

8 x + y^{2} + 4 y + 12 = 0

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

x, - 2 x + 4 k (x)^{2} = - 15

What is x^2-5x+4-M>= 0 ?
The solution to x^2-5x+4-M>= 0 is Parabola with (h,k)=(5/2 ,-9/4),p= 1/4