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

1 Answer
Dec 21, 2017

#"see explanation"#

Explanation:

#"given a parabola in "color(blue)"standard form"#

#•color(white)ax^2+bx+c color(white)(x);a!=0#

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

#•color(white)(x)x_(color(red)"vertex")=-b/(2a)#

#f(x)=2x^2+x-3" is in standard form"#

#"with "a=2, b=1, c=-3#

#rArrx_(color(red)"vertex")=-1/4#

#rArr"axis of symmetry is "x=-1/4#

#"substitute this value into the equation for y-coordinate"#

#y_(color(red)"vertex")=2(-1/4)^2+(-1/4)-3=-25/8#

#rArrcolor(magenta)"vertex "=(-1/4,-25/8)#

#"for "color(blue)"intercepts"#

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

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

#x=0toy=-3larrcolor(red)"y-intercept"#

#y=0to2x^2+x-3=0#

#rArr(x-1)(2x+3)=0#

#rArrx=1" and "x=-3/2larrcolor(red)"x-intercepts"#
graph{(y-2x^2-x+3)(y-1000x-250)=0 [-10, 10, -5, 5]}