what is a function with a vertex of (-3,-6) and passes through the point (-4,-3)

Answer 1

y= a(x+3)^2 -6

-3 = a(-4+3)^2 -6

-3 = a-6

a = 3

y=3(x+3)^2 -6 is an upward opening parabola

another relation is a leftward opening parabola, but it is not a function because there are 2 y values for any x value (except at the vertex). It fails the vertical line test.

x=a(y+6)^2 -3 where a<0

-4 = a(-3+6) -3

-4 = 3a -3

9a =-1

a = -1/9

x =(-1/9)(y+6)^2 -3

9x =- (y+6)^2 -27

9(-4) =- (-3+6)^2 -27

-36 = -9-27

Answer 2

Use the vertex form of a quadratic function:

y = a·(x-h)² + k

where a is a constant to be determined, and (h,k) is the location of the vertex. In your problem, (h,k) = (-3,-6), so h = -3 and k = -6:

y = a·(x-(-3))² + (-6)

y = a·(x+3)² - 6

To find the value of a, plug in the given point (x,y) = (-4, -3):

y = a·(x+3)² - 6

-3 = a·(-4+3)² - 6

Solve for a.