Question

How do you find the vertex, axis of symmetry and intercepts of `f(x)=3x^2+x-3`?

Answer 1

f(x) = 3x^2 + x - 3

Explanation:

x-coordinate of vertex, or axis of symmetry:
`x = -b/(2a) = -1/6`
y-coordinate of vertex:
`f(-1/6) = 3(1/36) - 1/6 - 3 = 1/12 - 1/6 - 3 = - 17/6`
To find y-intercept, make x = 0 --> y = -3
To find x-intercepts, make y = 0 and solve the quadratic equation
`y = 3x^2 + x - 3 = 0`
`D = d^2 = b^2 - 4ac = 1 + 36 = 37` -->`d = +- sqrt37`
There are 2 real roots, or two x-intercepts:
`x = -b/(2a) +- d/(2a) = -1/6 +- sqrt37/6`