Question

How do you find the vertex and intercepts for `f(x) = 3x^2 + 6x + 1`?

Answer 1

Vertex (-1,-2)
y-intercept (0,1)
x-intercept 1.8165, 0.1835

Explanation:

`y=a*(x-h)^2+k` is the equation of parabola, with vertex (h,k).
`y=3x^2+6x+1=3*(x^2+2x+1)-2`
`:.y=3*(x+1)^2-2`
Vertex (h,k) = `(-1,-2)`

To find y-intercept, let `x=0`
`y=3-2=1`;

To find x-intercept, let `y=0`
`0=3(x+1)^2-2`
`(x+1)^2=2/3`
`x=1-(+-sqrt(2/3))`
`x=1+sqrt(2/3), 1-sqrt(2/3) = 1.8165, 0.1835`