Find the vertex of the quadratic function. y=3(x+13)^2+1

Question

please help

Answer 1

The equation is already in the vertex form for a parabola, y=a(x-h)^2 + k. (h,k) is the vertex. In the problem, y=3 (x + 13)^2 + 1, h=-13 and k=1. The rule is to set the expression in the parentheses equal to 0 to find the x coordinate of the vertex, so x + 13 = 0 results in x or "h" = -13. For the y value of the vertex, "k" is the value of y as it appears in the equation so y or "k" = 1. The "a" value of positive 3 indicates that the parabola opens upward and is narrowed or stretched by a factor of 3.

Answer 2

For y = a(x - h)² + k, the vertex is at (h, k)

For y = 3(x - (-13))² + 1, the vertex is at (-13, 1)