Write the equation of a circle whose center is at the origin and contains the point (2, 3).

x^2+y^2=13

Step-by-step explanation:

radius is the distance from the origin to the point

r=(2^2+3^2)^(1/2)=13^(1/2)

x^2+y^2=(13^(1/2))^2=13

1. The vertex of the graph of the function is (-4, 7)

2. The vertex represents a minimum value

3. The equation represented by the new graph is g(x) = (x + 4)² + 1

Step-by-step explanation:

The given quadratic function is given in vertex  form, y = a·(x - h)² + k, as follows;

g(x) = (x + 4)² + 7

1. By comparing the given quadratic function and the vertex form of a quadratic equation, we have;

a = 1, h = -4, and k = 7

The vertex of the graph of the function, (h, k) = (-4, 7)

2. Given that a = 1 > 0, the graph of the quadratic function opens upwards and the vertex represents a minimum value

3. Shifting the graph 6 units down from where it is now will give;

The vertex = (h, k - 6) = (-4, 7 - 6) = (-4, 1)

h = -b/(2·a), k = (-b²/(4·a) + c = -a·h² + c

c = k + a·h²

Therefore, initial value of the constant term, c = -1×(-4)² + 7 = 23

After the shifting the graph 6 units down, c = 23 - 6 = 17

∴ k = 1 = -a·(-4)² + 17

∴ -16/16 = -1 = -a

a = 1

Therefore, the equation represented by the new graph is g(x) = (x + 4)² + 1.