Question

How do you find the intercepts and vertex for `f(x)= 3x^2+x-5`?

Answer 1

The `y` axis intercept is found by setting `x=0` in your equation:
The vertex is found considering your equation in the general quadratic form `y=ax^2+bx+c` so that the coordinates of the vertex are:
`x_v=-b/(2a)`
`y_v=-Delta/(4a)=-(b^2-4ac)/(4a)`
Yhe `x` axis interxept(s) are found (if they exist) by setting `y=0` and solving the second degree equation:
`3x^2+x-5=0` for `x`.