Question

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

Answer 1

vertex: `(-3/4,7/8)`
axis of symmetry: `x=-3/4`
intercept: `(0,2)`

Explanation:

Intercepts : `y` ints are where `x=0`, `x` ints where `y=0`. So to find the y intercept, set `x` to zero and solve:
`y=2(0)^2+3(0) +2`
`y=2`

`x` intercepts: set y to zero and solve (slightly trickier since the coefficient isn't one).
`2x^2+3x+2=0`
This doesn't have any real roots - the graph doesn't intercept the `x-"axis"`.

Vertex : if you don't know calculus, then this will be half way between the roots. Since this equation doesn't have any roots I'm just going to say that it's where the gradient is zero, which is when `4x+3=0` `=-3/4`
Y value: plug the x value into original equation: `2(-3/4)^2+3(-3/4)+2=0.875" or " 7/8`

Axis of symmetry:
This is also where the vertex is. Hence, axis of symmetry is `x=-3/4`

Here's the graph, which should make it all clearer.
graph{2x^2+3x+2 [-5, 5, -5, 10]}