Question 181670
{{{y=(x^2-x+1/4)+(5/4)-(1/4)}}}
:
{{{y=(x-1/2)^2+1}}}
:
vertex is ({{{highlight(1/2)}}},{{{1}}})
:
axis of symmetry is x=1/2
:
y intercept is found by setting x to zero y=0^2-0+5/4=5/4
:
y intercept is (0,5/4)
:
for b) and c) just plug in values of x and solve for y for as many points as you want
:
we already know that for x=0 y=5/4(the y intercept
:
we also know that for x=1/2 y=1
:
try more such as x=2--->y(2)^2-2+5/4=13/4
try some negative values, x=-1--->y=(-1)^2-(-1)+5/4=13/4
:
interesting both the same value but this makes sense, since vertex is at x=1/2
x=2 and x=-1 are equal distances from 1/2
:
hope that helps
:
oh here is the graph:
no x intercepts on this parabola
{{{graph(300,300,-10,10,-10,10,x^2-x+(5/4))}}}