Question

5 f(x)=(3x - 1)(3x + 1) dilate the function by x - 2. 6 g(x)=(x + 4)(x - 5) dilate the function by x + 3.

Explanation:

Step1: Recall dilation formula

If we dilate a function $y = f(x)$ by a factor $a$ about a point $x = c$, the new function is $y=\frac{1}{a}f(c + a(x - c))$. When dilating by $x - k$, we can rewrite it in terms of horizontal dilation. For dilating $y = f(x)$ by $x - 2$, let's first expand $f(x)=(3x - 1)(3x + 1)=9x^{2}-1$.

Step2: Apply dilation rule

The dilation of $y = f(x)$ by $x - 2$ means we replace $x$ with $\frac{x}{1}+2=x + 2$. So the new function $f_{new}(x)=9(x + 2)^{2}-1=9(x^{2}+4x + 4)-1=9x^{2}+36x+36 - 1=9x^{2}+36x + 35$.
For $g(x)=(x + 4)(x - 5)=x^{2}-x - 20$. When dilating by $x+3$, we replace $x$ with $\frac{x}{1}-3=x - 3$. So $g_{new}(x)=(x - 3+4)(x - 3 - 5)=(x + 1)(x - 8)=x^{2}-7x - 8$.

Answer:

For $f(x)=(3x - 1)(3x + 1)$, the dilated function by $x - 2$ is $y = 9x^{2}+36x + 35$. For $g(x)=(x + 4)(x - 5)$, the dilated function by $x + 3$ is $y=x^{2}-7x - 8$.