What are the #x# and #y#-intercepts of #h(x)=2x^2-x#?

2 Answers
May 11, 2018

#x_("intercept")=0#
#x_("intercept")=1/2#

Explanation:

Write as #y=2x^2-x+0#

#y_("intercept")="the constant"=0#

#x_("intercept")# is at #y=0# so set:

#y=0=2x^2-x#

#y=0=x(2x-1)#

So #x=0 and 2x-1=0#

#x_("intercept")=0#
#x_("intercept")=1/2#

May 11, 2018

#"x-intercepts "=0,1/2," y-intercept "=0#

Explanation:

#"to determine the x and y intercepts, that is where the"#
#"graph crosses the x and y axes"#

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

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

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

#y=0rArr2x^2-x=0#

#rArrx(2x-1)=0rArrx=0" or "x=1/2larrcolor(red)"x-intercepts"#
graph{2x^2-x [-10, 10, -5, 5]}