急急急 Quadradic equation (20分)

a) Since the curve opens upward, the first condition is a > 0

Also since the curve does not touch the x-axis, the discriminant of the equation:

ax2 + 2(a + 1)x + (a + 3) = 0

should be negative, i.e.

4(a + 1)2 - 4a(a + 3) < 0

(a + 1)2 - a(a + 3) < 0

a2 + 2a + 1 - a2 - 3a < 0

-a + 1 < 0

a > 1

b) Sub x = 4 into the curve function:

y = 16a + 8(a + 1) + (a + 3)

= 25a + 11

since a > 1, y > 36 and so y = 4 is impossible

(a)Since the curve opens upward,
so a>0
y = ax^2 + 2(a+1)x + a + 3
y=a[x^2+2(a+1)/a]+a+3
y=a{x^2+2(a+1)/a+[(a+1)/a]^2-[(a+1)/a]^2}+a+3
y=a[x+(a+1)/a]^2-(a^2+2a+1)/a+a+3
y=a[x+(a+1)/a]^2+(a-1)/a
the minimun point is ( -(a+1)/a , (a-1)/a )
the figure shown that
for all y, y>0
so (a-1)/a >0
a-1>0
a>1

(b)
assume point (-4,4) lies on the graph
so
-4=16a-8(a+1)+a+3
9a-1=0
a=1/9
it conflict with (a) a>1
so point(-4,4) does not lies on the graph

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http://www.mathopenref.com/quadraticexplorer.html