How do you find the axis of symmetry, vertex and x intercepts for #y=-x^2-3x+4#?

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Answer 1 · 2017-07-18T03:45:45

axis of symmetry is #x = -3/2#
vertex minimum at #25/4#
#x#-intercepts are #(1,0) and (-4,0)#

#y = - x^2 - 3 x + 4#

to find axis of symmetry and vertex, we can use completing a square to solve it.

#y = - (x^2 + 3 x) + 4#

#y = - (x + 3/2)^2 + (3/2)^2 + 4#

#y = - (x + 3/2)^2 + 9/4 + 16/4#

#y = - (x + 3/2)^2 + 25/4# #->i#

it axis of symmetry can be found when #(x + 3/2) = 0#, then #x = -3/2#

since #- (x + 3/2)^2# is always has a -ve value, therefore it has a maximum vertex at #25/4#

to find #x#-intercepts, plug in #y = 0# in #-> i#

#0 = - (x + 3/2)^2 + 25/4#

#(x + 3/2)^2 = 25/4#

#(x + 3/2) = +- sqrt (25/4)#

#x = - 3/2 +- 5/2#

#x_1 = - 3/2 + 5/2 = 2/2 = 1#

#x_2 = - 3/2 - 5/2 = -8/2 = -4#

therefore #x#-intercepts are #(1,0) and (-4,0)#