y = - x^2 - 3 x + 4y=−x2−3x+4
to find axis of symmetry and vertex, we can use completing a square to solve it.
y = - (x^2 + 3 x) + 4y=−(x2+3x)+4
y = - (x + 3/2)^2 + (3/2)^2 + 4y=−(x+32)2+(32)2+4
y = - (x + 3/2)^2 + 9/4 + 16/4y=−(x+32)2+94+164
y = - (x + 3/2)^2 + 25/4y=−(x+32)2+254 ->i→i
it axis of symmetry can be found when (x + 3/2) = 0(x+32)=0, then x = -3/2x=−32
since - (x + 3/2)^2−(x+32)2 is always has a -ve value, therefore it has a maximum vertex at 25/4254
to find xx-intercepts, plug in y = 0y=0 in -> i→i
0 = - (x + 3/2)^2 + 25/40=−(x+32)2+254
(x + 3/2)^2 = 25/4(x+32)2=254
(x + 3/2) = +- sqrt (25/4)(x+32)=±√254
x = - 3/2 +- 5/2x=−32±52
x_1 = - 3/2 + 5/2 = 2/2 = 1x1=−32+52=22=1
x_2 = - 3/2 - 5/2 = -8/2 = -4x2=−32−52=−82=−4
therefore xx-intercepts are (1,0) and (-4,0)(1,0)and(−4,0)
