find the vertex of y=(x+3)2+17

Answers

Answer 1

The vertex of the quadratic function [tex]y = (x + 3)^2 + 17[/tex] is (-3, 17).

This means that the parabola is symmetric around the vertical line x = -3 and has its lowest point at (-3, 17).

To find the vertex of the quadratic function y = (x + 3)^2 + 17, we can identify the vertex form of a quadratic equation, which is given by [tex]y = a(x - h)^2 + k,[/tex]

where (h, k) represents the vertex.

Comparing the given function [tex]y = (x + 3)^2 + 17[/tex]  with the vertex form, we can see that h = -3 and k = 17.

Therefore, the vertex of the quadratic function is (-3, 17).

To understand this conceptually, the vertex represents the point where the quadratic function reaches its minimum or maximum value.

In this case, since the coefficient of the [tex]x^2[/tex]  term is positive, the parabola opens upward, meaning that the vertex corresponds to the minimum point of the function.

By setting the derivative of the function to zero, we could also find the x-coordinate of the vertex.

However, in this case, it is not necessary since the equation is already in vertex.

For similar question on quadratic function.

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