Question

What is the vertex of the function?

`y=-3x^{2}+6x+15`

Answer 1

To find the vertex of the function, we first need to rewrite the function in the standard form of a quadratic equation: y = ax^2 + bx + c.

In this case, the function y = -3x^2 + 6x + 15 is already in standard form. Now we can identify the vertex of the parabola using the formula:

Vertex = (-b/2a, f(-b/2a))

In our function, a = -3 and b = 6.

Substitute these values into the formula:

Vertex = (-6 / 2(-3), f(-6 / 2(-3)))

Vertex = (-6 / -6, f(-6 / -6))

Vertex = (1, f(1))

To find f(1), substitute x = 1 into the original function:

y = -3(1)^2 + 6(1) + 15
y = -3 + 6 + 15
y = 18

Therefore, the vertex of the function y = -3x^2 + 6x + 15 is (1, 18).