Question

use transformations of $f(x) = \frac{1}{x^2}$ to graph $g(x) = \frac{1}{(x + 2)^2} - 7$. select the correct graph. \\(\bigcirc\\) a. graph a \\(\bigcirc\\) b. graph b \\(\bigcirc\\) c. graph c \\(\bigcirc\\) d. graph d

Explanation:

Step1: Analyze horizontal transformation

For the function \(y = f(x + h)\), when \(h>0\), the graph of \(y = f(x)\) is shifted \(h\) units to the left.
For \(g(x)=\frac{1}{(x + 2)^{2}}\), compared to \(f(x)=\frac{1}{x^{2}}\), here \(h = 2\). So the graph of \(y=\frac{1}{x^{2}}\) is shifted 2 units to the left.

Step2: Analyze vertical transformation

For the function \(y=f(x)+k\), when \(k<0\), the graph of \(y = f(x)\) is shifted \(|k|\) units down.
For \(g(x)=\frac{1}{(x + 2)^{2}}-7\), compared to \(y=\frac{1}{(x + 2)^{2}}\), here \(k=-7\). So the graph of \(y=\frac{1}{(x + 2)^{2}}\) is shifted 7 units down.

The vertex of \(f(x)=\frac{1}{x^{2}}\) is \((0,0)\). After the transformation:

• The horizontal shift: the \(x\) - coordinate of the vertex changes from \(x = 0\) to \(x=-2\) (because of \(x+2\) in the denominator).
• The vertical shift: the \(y\) - coordinate of the vertex changes from \(y = 0\) to \(y=-7\)

Answer:

Assuming the graph with the vertex at \((-2,-7)\) (after applying left - shift by 2 units and down - shift by 7 units) is the correct one. Without seeing the exact visual details of each option (but based on transformation rules), if we consider the vertex position:
If option D has the vertex at \((-2,-7)\) (after shifting the basic \(y=\frac{1}{x^{2}}\) graph 2 units left and 7 units down), then D is the answer.