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

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 $y = a {x}^{2} + b x + c$ so that the coordinates of the vertex are:
${x}_{v} = - \frac{b}{2 a}$
${y}_{v} = - \frac{\Delta}{4 a} = - \frac{{b}^{2} - 4 a c}{4 a}$
Yhe $x$ axis interxept(s) are found (if they exist) by setting $y = 0$ and solving the second degree equation:
$3 {x}^{2} + x - 5 = 0$ for $x$.