How do you find the intercept and vertex of #y= (2x-1)(3x+4)#?

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Answer 1 · 2016-03-06T00:16:44

Analyse the equation to find #y# intercept #(0, -4)#, #x# intercepts #(-4/3, 0)#, #(1/2, 0)# and vertex #(-5/12, -121/24)#

If #y=0# then #(2x-1) = 0# or #(3x+4) = 0#

Hence #x=1/2# or #x=-4/3#

If #x=0# then #y = (0-1)(0+4) = -4#

So the #x# intercepts are #(-4/3, 0)# and #(1/2, 0)#

and the #y# intercept is #(0, -4)#

Since parabolas are bilaterally symmetric, the vertex will have #x# coordinate exactly midway between the two #x# intercepts:

#x = (-4/3+1/2)/2 = (-8/6+3/6)/2 = -5/12#

Substitute this value of #x# back into the original equation to find:

#y = (2(-5/12)-1)(3(-5/12)+4)#

#=(-5/6-6/6)(-15/12+48/12)#

#=(-11/6)(33/12)#

#=(-11/6)(11/4)#

#=-121/24#

So the vertex is at #(-5/12, -121/24)#

graph{(2x-1)(3x+4) [-6.58, 5.904, -5.27, 0.97]}