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