Question

For which pair of functions is the vertex of k(x)7 units below the vertex of f(x)?

Answer 1

Answer: Option C

$f(x) = x^2;\ k (x) = x ^ 2 -7$

Step-by-step explanation:

Whenever we have a main function f(x) and we want to transform the graph of f(x) by moving it vertically then we apply the transformation:

$k (x) = f (x) + b$

If $b> 0$ then the graph of k(x) will be the graph of f(x) displaced vertically b units down.

If $b> 0$ then the graph of k(x) will be the graph of f(x) displaced vertically b units upwards.

In this case we have

$f (x) = x ^ 2$

We know that this function has its vertex in point (0,0).

Then, to move its vertex 7 units down we apply the transformation:

$k (x) = f (x) - 7\\\\k (x) = x ^ 2 -7$.

Then the function k(x) that will have its vertex 7 units below f(x) is

$k (x) = x ^ 2 -7$