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

1 Answer
Gió
Apr 20, 2015

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#.

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