How do you find the vertex, axis of symmetry and intercepts of f(x)=2x2+x=3?
1 Answer
Dec 21, 2017
Explanation:
given a parabola in standard form
∙ax2+bx+cx;a≠0
then the x-coordinate of the vertex which is also the
axis of symmetry is
∙xxvertex=−b2a
f(x)=2x2+x−3 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=0→y=−3←y-intercept
y=0→2x2+x−3=0
⇒(x−1)(2x+3)=0
⇒x=1 and x=−32←x-intercepts
graph{(y-2x^2-x+3)(y-1000x-250)=0 [-10, 10, -5, 5]}