How do you find the vertex and intercepts for #y=2x^2 +11x-6#?

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Answer 1 · 2017-09-14T12:47:13

Vertex is at # (-2.75, -21.125) # x intercepts are at
# (-6,0) and (0.5,0)#, y intercept is at # (0, -6)#

#y= 2x^2+11x-6 or y= 2(x^2+11/2x)-6 #

Adding #121/8# on both sides we get ,

#y= 2{x^2+11/2x+(11/4)^2] -121/8 -6 # or

#y= 2(x+11/4)^2 -169/8 # or

#y= 2(x+2.75)^2 -21.125 # . Comparing with standard vertex form

of equation #y=a(x-h)^2+k ; (h,k)# being vertex , we find here

#h= -2.75 , k = -21.125 , a=2#. So vertex is at # (h,k) #or

# (-2.75, -21.125) # , y intercept can be found by putting

#x=0# in the equation as # y = 2*0 +11*0 -6# or

#y = -6 or (0,-6)# , x intercept can be found by putting #y=0#

in the equation as # 0 = 2x^2 +11x -6 or 2x^2 +12x -x -6=0 #

or # 2x(x+6) -1 (x+6) =0 or (2x-1)(x+6)=0#

# :. x = 0.5 or x= -6 #, x intercepts are at # (-6,0) and (0.5,0)#

graph{2x^2+11x-6 [-80, 80, -40, 40]} [Ans]