Functions, Help?

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1 Answer
May 4, 2018

(i) #x<3/2 or x>3#

(ii) #(3/4,-9/8)#

(iii) #k=27/8#

Explanation:

(i)

#y=2x^2-3x#

When #y>9#,

#2x^2-3x>9#

Subtract #9# from both sides,

#2x^2-3x-9>0#

Factor,

#(2x+3)(x-3)>0#

Graph,

graph{2x^2-3x-9 [-5, 5, -15, 15]}

From the graph,

#x<3/2 or x>3#

(ii)

#2x^2-3x#

Factor,

#2(x^2-3/2x)#

Perform complete the square method,

Let us recap our concepts,

#color(red)(a^2)+color(green)(2ab)+color(blue)(b^2)=color(purple)((a+b)^2#

Let #a=x# and #2ab=-3/2x#

#b=-(3/2x)/(2x)#
#color(white)(b)=-3/4#

Now we have to add in #color(blue)((-3/4)^2)# somehow,

#2(color(red)(x^2)-color(green)(3/2x)+color(blue)((-3/4)^2)-(-3/4)^2)#

By adding and subtracting, I have not affected the original equation,

#2(color(red)(x^2)-color(green)(3/2x)+cancel(color(blue)((-3/4)^2))-cancel((-3/4)^2))#
#=2(x^2-3/2x)#

Now simplify,

#2(color(red)(x^2)-color(green)(3/2x)+color(blue)((-3/4)^2))-2(-3/4)^2#

Make the perfect square and simplify,

#2color(purple)((x-3/4)^2)-9/8#

Hence,

#2(x-3/4)^2-9/8# - ( vertex form )

The vertex formula is as follows:

#a(x-h)^2+k#, where #(h,k)# is the vertex.

By comparison, find the coordinates,

#(3/4,-9/8)#

(iii)

#f(x)=2x^2-3x#

#g(x)=3x+k#

Substitute,

#gf(x)=g(2x^2-3x)#
#color(white)(gf(x))=3(2x^2-3x)+k#

When #gf(x)=0#,

#3(2x^2-3x)+k=0#

Expand,

#6x^2-9x+k=0#

Apply discriminant, where #b^2-4ac=0# for equal roots.

#(-9)^2-4(6)(k)=0#

Simplify,

#81-24k=0#

Add #24k# to both sides,

#24k=81#

Solve,

#k=81/24#
#color(white)(k)=27/8#