How do you find the vertex, axis of symmetry and intercepts of f(x)=2x2+3x+2?

1 Answer
TreeReader
Nov 19, 2017

vertex: (−34,78)
axis of symmetry: x=−34
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).
2x2+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 =−34
Y value: plug the x value into original equation: 2(−34)2+3(−34)+2=0.875 or 78

Axis of symmetry:
This is also where the vertex is. Hence, axis of symmetry is x=−34

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

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
2021 views around the world
You can reuse this answer
Creative Commons License