Given the function:
y=2x2+6
Determine for which values of x is f\left(x\right) > 0 true
To solve this problem, let's analyze the quadratic function y=2x2+6:
- The function y=2x2+6 is a parabola opening upwards because the coefficient of x2 is positive (a=2).
- A parabola that opens upwards has its vertex as the lowest point.
- Since the function is in the form y=ax2+c with no x term, there are no real roots (our function has no b term, hence no rotations impacting x-intercepts).
- We evaluate f(x)>0: Since the constant term c=6 is positive, this defines a vertical shift of the parabola entirely above the x-axis.
- This implies y=2x2+6 is always greater than zero for any value of x.
Therefore, the solution to the problem is all x.