Question 162673
See this for a fairly detailed look at this issue --> http://www.analyzemath.com/quadratics/vertex_problems.html

I assume you meant {{{3x^2+x-5}}}
Any point that is an x axis intercept must have a y coordinate value of 0. So to solve for those points. solve
{{{3x^2+x-5=0}}} You can use quadratic equation for that.
*[invoke quadratic "x", 3, 1, -5]

The URl above gives you a formula for finding the vertex. Here are two others ways
1)if you know deriviatives, take the first derivative and set that to 0. Then solve for x. Use that x vlaue to find the corresponding y in the original equation.
2) find the x value that is the average of the 2 x intercepts. That point is on the line that 'splits' the parabola (the line about which the parabola has symmetry). Find that x value and plug it into the original equation to find the y value.