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Answer:
- y'=10x+7; y''=10; vertex: (-0.7, 0.55), min; no real zeros
- y'=-2x-5; y''=-2; vertex: (-2.5, 8.75), max; zeros: {-5.458, 0.458}
Explanation:
For a general parabola ...
y = ax² +bx +c
The first derivative is ...
y' = 2ax +b
and the second derivative is ...
y'' = 2a
The second derivative is a (non-zero) constant, so there is no point of inflection. The sign of the second derivative (the sign of 'a') tells you whether the graph opens upward (a>0) or downward (a<0).
The first derivative is 0 where ...
y' = 0 = 2ax +b
x = -b/(2a) . . . . . extreme point of the graph
Whether this point is a maximum (a<0) or a minimum (a>0) depends on the sign of 'a'.
The value of the function at the extreme is ...
y = a(-b/2a)² +b(-b/2a) +c = b²/(4a) -b²/(2a) +c
y = c -b²/(4a)
So, the extremum is ...
(x, y) = (-b/(2a), c -b²/(4a)) . . . . . vertex of the parabola
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1. We have a=5, b=7, c=3.
First derivative: f'(x) = 10x +7
Second derivative: f''(x) = 10 . . . . graph opens upward
Zeros: the discriminant b² -4ac is 7²-4(5)(3) = -11, so no real zeros
Vertex: (-7/10, 3 -49/20) = (-0.7, 0.55) . . . minimum
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2. We have a=-1, b=-5, c=5/2.
First derivative: f'(x) = -2x -5
Second derivative: f''(x) = -2 . . . . graph opens downward
Zeros: found from the quadratic formula: x = (-b±√(b²-4ac))/(2a)
x = (-(-5) ±√((-5)² -4(-1)(5/2)))/(2(-1)) = (5 ±√35)/2 ≈ {-5.458, 0.458}
Vertex: (-5/2, 5/2-25/-4) = (-5/2, 35/4) = (-2.5, 8.75) . . . maximum