In order to find the vertex of this quadratic equation, first let's find the coefficients a, b and c from the standard form of the quadratic equation:
[tex]y=ax^2+bx+c[/tex]
Comparing with the given equation, we have a = -5, b = -270 and c = -520.
Now, let's calculate the x-coordinate of the vertex using the formula below:
[tex]\begin{gathered} x_v=\frac{-b}{2a} \\ x_v=\frac{-(-270)}{2\cdot(-5)} \\ x_v=\frac{270}{-10} \\ x_v=-27 \end{gathered}[/tex]
Using this value of x in the equation, we can find the y-coordinate of the vertex:
[tex]\begin{gathered} y_v=-5x^2_v-270x_v-520 \\ y_v=-5\cdot(-27)^2-270\cdot(-27)-520 \\ y_v=-5\cdot729+7290-520 \\ y_v=-3645+7290-520 \\ y_v=3125 \end{gathered}[/tex]
Therefore the vertex is located at (-27, 3125).
