The red curve [latex]g(x)[/latex] appears to be less steep compared to the green curve [latex]f(x)[/latex]. This suggests a vertical compression.If [latex]g(x)[/latex] is a vertical compression of [latex]f(x)[/latex], we have: [latex]g(x) = a \cdot f(x)[/latex], where [latex]0 < a < 1[/latex].To determine [latex]a[/latex], it is helpful to look for a point on the graph that is relatively clear.

• In this graph, it appears that [latex]g\left(2\right)=2[/latex].
• With the basic cubic function at the same input, [latex]f\left(2\right)={2}^{3}=8[/latex].
• Based on that, it appears that the outputs of [latex]g[/latex] are [latex]\frac{1}{4}[/latex] the outputs of the function [latex]f[/latex] because [latex]2=\frac{1}{4} \cdot 8[/latex].

Thus, [latex]g(x) = \frac{1}{4} f(x) = \frac{1}{4} x^3[/latex].

The orange graph  [latex]g(x)[/latex] appears to be a horizontally compressed version of the blue graph of [latex]f(x)[/latex].
[latex]\\[/latex]
If [latex]g(x)[/latex] is a horizontal compression of [latex]f(x)[/latex], we have: [latex]g(x) = f(b \cdot x)[/latex], where [latex]b > 1[/latex]. The graph is compressed by [latex]\dfrac{1}{b}[/latex].To determine [latex]b[/latex], it is helpful to look for a point on the graph that is relatively clear.

• In the compressed graph [latex]g(x)[/latex], the end point is [latex](2, 4)[/latex].
• The end point of [latex]f(x)[/latex] is [latex](6,4)[/latex].
• We can see that the [latex]x[/latex]-values have been compressed by [latex]\frac{1}{3}[/latex], because [latex]2=\frac{1}{3} \cdot 6[/latex].
• This means that [latex]\dfrac{1}{b} = \dfrac{1}{3}[/latex], which means [latex]b = 3[/latex].

Thus, [latex]g(x)=f(3x)[/latex].

Notice that the graph is identical in shape to the [latex]f\left(x\right)={x}^{2}[/latex] function, but the [latex]x[/latex]–values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so

[latex]g\left(x\right)=f\left(x - 2\right)[/latex]

Notice how we must input the value [latex]x=2[/latex] to get the output value [latex]y=0[/latex]; the [latex]x[/latex]-values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the [latex]f\left(x\right)[/latex] function to write a formula for [latex]g\left(x\right)[/latex] by evaluating [latex]f\left(x - 2\right)[/latex].

[latex]\begin{cases}f\left(x\right)={x}^{2}\hfill \\ g\left(x\right)=f\left(x - 2\right)\hfill \\ g\left(x\right)=f\left(x - 2\right)={\left(x - 2\right)}^{2}\hfill \end{cases}[/latex]

Analysis of the Solution

To determine whether the shift is [latex]+2[/latex] or [latex]-2[/latex] , consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, [latex]f\left(0\right)=0[/latex]. In our shifted function, [latex]g\left(2\right)=0[/latex]. To obtain the output value of 0 from the function [latex]f[/latex], we need to decide whether a plus or a minus sign will work to satisfy [latex]g\left(2\right)=f\left(x - 2\right)=f\left(0\right)=0[/latex]. For this to work, we will need to subtract 2 units from our input values.

The graph of the toolkit function starts at the origin, so this graph has been shifted 1 to the right and up 2. In function notation, we could write that as

[latex]h\left(x\right)=f\left(x - 1\right)+2[/latex]

Using the formula for the square root function, we can write

[latex]h\left(x\right)=\sqrt{x - 1}+2[/latex]

Analysis of the Solution

Note that this transformation has changed the domain and range of the function. This new graph has domain [latex]\left[1,\infty \right)[/latex] and range [latex]\left[2,\infty \right)[/latex].