The equation y=3x^2+2 is a quadratic equation in the form of y=ax^2+bx+c. In this equation, a=3, b=0, and c=2.
The graph of this equation is a parabola that opens upwards since the coefficient of the x^2 term (a) is positive. The vertex of the parabola is located at the point (-b/2a, c).
In this case, the vertex can be found by substituting the values of a, b, and c into the formula for the x-coordinate of the vertex: x = -b/2a.
x = -0/2(3)
x = 0
Therefore, the vertex is located at the point (0, 2).
The graph of this equation will intersect the y-axis at y=c, which is 2 in this case. Thus, the y-intercept is at (0, 2).
To find the x-intercepts, we need to solve the equation for x when y=0.
0 = 3x^2 + 2
Subtracting 2 from both sides, we get:
-2 = 3x^2
Dividing both sides by 3, we get:
x^2 = -2/3
Since there is no real number that satisfies this equation, there are no x-intercepts for this quadratic equation.
The graph of y=3x^2+2 is a parabola that opens upwards, with a vertex at (0, 2), a y-intercept at (0, 2), and no x-intercepts.
y=3x^2+2
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