Answer:
[tex]y = 5x{^2} - 30x + 39[/tex]
Step-by-step explanation:
Given the vertex form of the equation of a parabola to be [tex]y = 5(x-3)^{2} - 6[/tex]
The standard form of the equation will be a quadratic equation in the form;
[tex]y = ax^{2} + bx + c[/tex]
where,
y is dependent variable
x is independent variable
a and b are constant coefficients of independent variable x² and x respectively
c is a constant
Transforming the vertex form to the standard form of a quadratic function y, we develop the equation:
[tex]y = 5(x-3)^{2} - 6[/tex]
[tex]y = 5(x-3)(x-3) - 6[/tex]
[tex]y = 5(x^{2} - 6x + 9) - 6[/tex]
[tex]y = 5x{^2} - 30x + 45 - 6[/tex]
[tex]y = 5x{^2} - 30x + 39[/tex]
The standard form of the equation is [tex]y = 5x{^2} - 30x + 39[/tex]
