the vertex of a parabola is located at (-2,5). If the parabola has a y-intercept of -27, which quadratic function represents the parabola in standard form? y=a(x-p)^2 +q

Step-by-step explanation:

y = a(x - p)² + q  - the vertex form of an equation of a parabola with a vertex (p, q)

(-2, 5)  ⇒  p = -2,   q = 5

y = a(x + 2)² + 5   - the vertex form of an equation of the parabola with vertex (-2, 5)

y-intercept of -27 means x=0,  y=-27

-27 = a(0 + 2)² + 5

-27 -5 = a(2)² + 5 -5

-32 = 4a

  a = -8

The equation of the parabola in vertex form that has a vertex (-2, 5) and y-intercept of -27:

                              f(x) = -8(x + 2)² + 5

i believe it’s 165 because if u add 137 and 58 u get 195 and then subtract that from 360 and there’s ur answer