Question

Question 8 of 10 for which pair of functions is the vertex of k(x) 7 units below the vertex of f(x)?

a. f(x) = x² and k(x) = (x – 7)²
b. f(x) = x² and k(x) = x² – 7
c. f(x) = x² and k(x) = x² + 7
d. f(x) = x² and k(x) = (x + 7)²

Answer 1

The correct pair of functions where the vertex of k(x) is 7 units below the vertex of f(x) is b. f(x) = x^2 and k(x) = x^2 - 7, as this reflects a downward shift in the vertex by 7 units.

Step-by-step explanation:

To determine which function k(x) has its vertex 7 units below that of function f(x), it is important to understand the effects of transformations on the graph of a quadratic equation. The vertex form of a quadratic function is f(x) = a(x - h)^2 + k, where (h, k) is the vertex of the parabola. A shift upwards or downwards is reflected by the + k value.

Considering f(x) = x^2, its vertex is at (0, 0). To move the vertex down by 7 units, we should subtract 7 from the function, resulting in k(x) = x^2 - 7. This tells us that the correct answer is:

• b. f(x) = x^2 and k(x) = x^2 - 7