Question

suppose ( g(x) ) is a transformation of ( f(x) ) such that ( f(x) ) translates left 11 units, stretches vertically by a factor of ( \frac{3}{4} ) and translates up 7 units. enter the function ( g(x) ).

Explanation:

Step1: Recall function transformation rules

For a function \(y = f(x)\), a horizontal translation \(a\) units to the left gives \(y = f(x + a)\), a vertical stretch by a factor \(b\) gives \(y=b\cdot f(x)\), and a vertical translation \(c\) units up gives \(y=f(x)+c\).

Step2: Analyze the given transformations

We know \(f(x)\) is translated left \(11\) units, so we replace \(x\) with \(x + 11\) in \(f(x)\) (i.e., \(f(x)\to f(x + 11)\)). Then it is vertically stretched by a factor of \(\frac{3}{4}\), so we get \(\frac{3}{4}f(x + 11)\). Finally, it is translated up \(7\) units, so we add \(7\) to the function.

Answer:

\(g(x)=\frac{3}{4}f(x + 11)+7\)