Which quadratic function in vertex form has a vertex at (−5 , −4) and passes through the point (−2 , −1)? y=−13(x+5)^2−4

y=−13(x−5)^2+4

y=13(x−5)^2−4

y=13(x+5)^2−4

y = 1/3(x + 5)^2 - 4

Step-by-step explanation:

The general vertex form is;

y = a(x-h)^2 + k

From the question, (h,k) are the coordinates of the vertex which is (-5,-4)

So we have

y = a(x + 5)^2 -4

So we need to get the leading coefficient

Simply substitute the point it passes through

x = -2 and y = -1

-1 = a(-2 + 5)^2 - 4

-1 = a(3)^2 -4

-1 = 9a -4

9a = 4-1

9a = 3

a = 3/9

a = 1/3

So the equation is;

y = 1/3(x + 5)^2 - 4

7

step-by-step explanation:

the answer would be 7

i believe that the slope of the line is -5

step-by-step explanation: