vertex a minimum or maximum?
The equation in vertex form y=3 x^2+4
The vertex of the parabola represented by the equation y=3x^2+4 is a minimum point.
To see why, recall that the vertex form of a quadratic equation is y = a(x-h)^2 + k, where (h, k) is the vertex of the parabola.
In this case, a = 3, which means the parabola opens upward. Since a is positive, the vertex is a minimum point.
To find the coordinates of the vertex, we can use the fact that the vertex occurs at x = -b/(2a), where in this case b = 0 since there is no x-term. Therefore, x = 0. Substituting this into the equation, we get:
y = 3(0)^2 + 4 = 4
So the coordinates of the vertex are (0, 4), and the parabola opens upward with a minimum value of 4.
Note that as x goes big + or - , y goes straight up :)
Yes, that's correct! As x approaches positive or negative infinity, the value of y increases without bound, since the parabola opens upward.
To determine whether the given equation is a minimum or maximum, we need to look at the coefficient of the x^2 term, which is positive in this case.
Since the coefficient is positive (3), the parabola opens upwards, indicating that the vertex is a minimum.
Therefore, the given equation has a minimum.