A quadratic equation, \( \mathrm{f}: x \mapsto a x^{2}+b x+c \), has a vertex at \( (-2,27) \) and passes through the \( y \)-axis at \( y=15 \). Find the equation and state the values of \( a \) and \( c \) separated by a comma and with no spaces, e.g.: \( -15,2 \).

To find the quadratic equation \( f(x) = ax^2 + bx + c \) with the given conditions, we start with the vertex form of a quadratic equation: \[ f(x) = a(x - h)^2 + k \] where \( (h, k) \) is the vertex. Given that the vertex is at \( (-2, 27) \), we have: \[ h = -2, \quad k = 27 \] Thus, the equation can be written as: \[ f(x) = a(x + 2)^2 + 27 \] Next, we know that the quadratic passes through the \( y \)-axis at \( y = 15 \). This means when \( x = 0 \), \( f(0) = 15 \). We can substitute \( x = 0 \) into the equation: \[ f(0) = a(0 + 2)^2 + 27 = 15 \] This simplifies to: \[ f(0) = a(2^2) + 27 = 15 \] \[ 4a + 27 = 15 \] Now, we solve for \( a \): \[ 4a = 15 - 27 \] \[ 4a = -12 \] \[ a = -3 \] Now that we have \( a \), we can substitute it back into the vertex form to find \( c \). The standard form of the quadratic equation is: \[ f(x) = -3(x + 2)^2 + 27 \] Expanding this: \[ f(x) = -3(x^2 + 4x + 4) + 27 \] \[ f(x) = -3x^2 - 12x - 12 + 27 \] \[ f(x) = -3x^2 - 12x + 15 \] From this, we can identify the coefficients: \[ a = -3, \quad c = 15 \] Thus, the values of \( a \) and \( c \) are: \[ \boxed{-3,15} \]