(a) a = 1, b = 2, c = 3
\(\therefore\) D = b2 - 4ac = 4 - 12 = -8 < 0
No real roots
Vertex \((\frac{-b}{2a},\frac{-D}{4a})=(-1,(\frac{-8}4)) = (-1, +2)\)
(b) a = -4, b = -6, c = 2
\(\therefore\) D =b2 - 4ac = 36 + 32 = 68 > 0
two different real roots.
root are x = \(\frac{-b\pm\sqrt D}{2a}\) = \(\frac{6\pm\sqrt{68}}{-8}\)
one positive and one negative root.(\(\because\sqrt{68}>8\))
vertex \((\frac{-b}{2a},\frac{-D}{4a})=(\frac6{-8},\frac{-68}{-16})\) = \((\frac{-3}{4},\frac{17}4)\)
(c) a > 0, b < 0, c > 0, b2 - 4ac < 0
\(\therefore\) No real roots
\(\therefore\) \(\frac{-b}{2a} >0\) & \(\frac{-D}{4a} > 0\)
(d) a = c, b = 2, D = 0
two equal real roots.
\(\therefore\) x2 = \(\frac{c}a=1\)
⇒ x = 1 or x = -1
vertex (\(\frac{-b}{2a},\frac{-D}{4a}\)) = \((\frac{-2}{2a},0)=(\frac{-1}a,0)\)
Case I: a > 0 then c >0 (\(\therefore\) y - intercept is positive)
Case II: a < 0 then c < 0 (y - intercept is negative)
