We are given the function f(x) = x^3 - 6x^2 + 10x - 3, where x is a real number. We are supposed to translate it by (-2, 3) which means that we are moving the entore graph 2 units to the right and 3 units up.
The first thing to do is write the cubic function in the form f(x) = a(x - h)^3 + k where h is the horizontal translation and k is the vertical translation.
So, f(x) = x^3 - 6x^2 + 10x - 3 moved 2 units to the right and 3 units up would be written as:
[tex]C=(x+2)^3-6(x+2)^2+10(x+2)-3+3[/tex]
We can simplify this as:
[tex]\begin{gathered} C=(x+2)^{3}-6(x+2)^{2}+10(x+2)-3+3 \\ C=(x^3+6x^2+12x+8)-6(x^2+4x+4)+(10x+20) \\ C=x^3+6x^2+12x+8-6x^2-24x-24+10x+20 \\ C=x^3-2x+4 \end{gathered}[/tex]
The graph crosses the y-axis when x = 0.
[tex]\begin{gathered} C=x^{3}-2x+4 \\ y=0^3-2(0)+4 \\ y=4 \end{gathered}[/tex]
The point where the graph crosses the y-axis is (0, 4).
