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

1 Answer
Dec 21, 2017

see explanation

Explanation:

given a parabola in standard form

ax2+bx+cx;a0

then the x-coordinate of the vertex which is also the
axis of symmetry is

xxvertex=b2a

f(x)=2x2+x3 is in standard form

with a=2,b=1,c=3

xvertex=14

axis of symmetry is x=14

substitute this value into the equation for y-coordinate

yvertex=2(14)2+(14)3=258

vertex =(14,258)

for intercepts

let x = 0, in the equation for y-intercept

let y = 0, in the equation for x-intercepts

x=0y=3y-intercept

y=02x2+x3=0

(x1)(2x+3)=0

x=1 and x=32x-intercepts
graph{(y-2x^2-x+3)(y-1000x-250)=0 [-10, 10, -5, 5]}