Can someone please tell me how to find the vertex for y=x^2-7
Thanks!
Wouldn't the mininum y be at x=0 (y=-7?)
sure, I don't know, lol
To find the vertex of a quadratic function in the form y = ax^2 + bx + c, you can use the formula x = -b/2a to find the x-coordinate of the vertex.
In the equation y = x^2 - 7, we can see that a = 1, b = 0, and c = -7. Plugging these values into the formula, we get:
x = -b / (2a) = -0 / (2 * 1) = 0
So the x-coordinate of the vertex is x = 0.
To find the y-coordinate of the vertex, you can substitute the value of x into the original equation.
y = (0)^2 - 7 = 0 - 7 = -7
Therefore, the vertex of the quadratic function y = x^2 - 7 is (0, -7).
In this particular case, you are correct that the minimum value of y is at x = 0, with y = -7. The vertex represents the point on the graph where the parabola reaches its maximum or minimum value. In this case, since the coefficient of x^2 is positive (a = 1), the parabola opens upward and the vertex represents a minimum point.