Question

The point (2,8) is on the graph of f(x)=x3. What is the value of b if the point (0.5,8) is on the graph of f(bx)?

Answer 1

1) Given $f(x) = x^{3}$

              $f(bx) = (bx)^{3}= b^{3} x^{3}$

Since (0.5,8) lies on f(bx) , $8=b^{3} 0.5^{3}$

    Divide with $0.5^{3}$ on both sides

                                           $b^{3} = \frac{8}{0.5^{3} }$

                                                           = 64

                                                   $b=64^{\frac{1}{3} } = 4$

Hence b=4.

2) Given (1,3) lies on y=f(x) plugin x=1 and y=3

that is f(1) = 3

We can find corresponding point for that in y=-2f(x) by plugging in x=1.

That is y= -2f(1) = -2*3 = -6

Hence point is (1,-6)