Question

The vertex form of a quadratic function is f(x) = a(x - h)2 + k. What is the vertex of each function? Match the function rule with the coordinates of its vertex.

Answer 1

The answer is (vertex (5,-9) and using Tex

Answer 2

Answer:

For  $f(x)=6(x-5)^2-9$   the vertex is: (5, -9)

For  $f(x)=9(x+5)^2-6$  the vertex is: (-5, -6)

For  $f(x)=5(x-6)^2+9$  the vertex is: (6, 9)

For  $f(x)=6(x+9)^2-5$  the vertex is: (-9, -5)

For  $f(x)=9(x-5)^2+6$  the vertex is: (5, 6)

Step-by-step explanation:

Let's identify the vertex pair $(x_v,y_v)$ from each equation:

A)  $f(x)=6(x-5)^2-9$  corresponds to $x_v=5\,\,\,and\,\,\,y_v=-9$ , that is: (5, -9)

B)  $f(x)=9(x+5)^2-6$  corresponds to $x_v=-5\,\,\,and\,\,\,y_v=-6$ , that is: (-5, -6)

C)  $f(x)=5(x-6)^2+9$  corresponds to $x_v=6\,\,\,and\,\,\,y_v=9$ , that is: (6, 9)

D)  $f(x)=6(x+9)^2-5$  corresponds to $x_v=-9\,\,\,and\,\,\,y_v=-5$ , that is: (-9, -5)

E)  $f(x)=9(x-5)^2+6$  corresponds to $x_v=5\,\,\,and\,\,\,y_v=6$ , that is: (5, 6)