Question

find the interval of decreasing y=-x^2+4x+3

Answer 1

f(x) = -x² + 4x + 3|
$\frac{d}{dx}[-x^{2}] + \frac{d}{dx}[4x] + \frac{d}{dx}[3]|$
$-2x + 4 + 0|$
$-2x + 4|$

$\frac{d}{dx}[x^{n}]| = nx^{n - 1}|$
$\frac{d}{dx}[-x^{2}]| = 2(-x)^{2 - 1}|$
$\frac{d}{dx}[-x^{2}| = 2(-x)^{1}|$
$\frac{d}{dx}[-x^{2}]| = 2(-x)|$
$\frac{d}{dx}[-x^{2}]| = -2x|$

$\frac{d}{dx}[4x]|$
$\frac{4dx}{dx}|$
$4(1)|$
$4|$

$\frac{d}{dx}[3]|$
$\frac{3d}{dx}|$
$3(1)(0)|$
$3(0)|$
$0|$

-2x + 4 = 0|
       - 4  - 4
      -2x = -4|
       -2     -2
         x = 2|

(-∞, 2) ∨ (2, ∞)|
(-∞, 2)| ∨| (2, ∞)|

(-∞, 2)
f'(x) = -2x + 4|
f'(2) = -2(2) + 4|
f'(2) = -4 + 4|
f'(2) = 0|

At 2| the derivative is 0.| Since this is positive, the function is increasing on (-∞, 2).
Increasing On: (-∞,| 2)| since f(x)| > 0|

(2, ∞)
f'(x) = -2x + 4|
f'(2) = -2(2) + 4|
f'(2) = -4 + 4|
f'(2) = 0|

At 2| the derivative is 0.| Since this is negative, the function is decreasing on (2, ∞)|.
Decreasing On: (2,| ∞)| since f(x)| < 0|

THe interval of the decreasing function y = -x² + 4x + 3 is (2, 2), or 2 > x > 2 and 2 < x < 2.