hi dude, for this question you have to learn how to draw parabola from quadratic equation, answer is asking you curve not line!
the answer is true, it doesnt intersecet quadrant II, but how?
y=-(3-x)^2+4=-x^2+6x-5
when y=o, then x1=1, x2=5 where the curve intersect x-axis:
when x=0, then y=-5 this where one side of curve intersects y-axis.
now we have to find peak point point of parabola which is equal to -b/2a from ax^2+bx+c=0, here b=6 and a =-1 then peak point is 3 from y point, the maximum or minium point of curve.
x1=1, x=5
y=-5
peak point is y=3
now try to draw the parabola,
it begins from y=3
left side intersects (1,0) and continues till intersects again (0,-5)
right side intersects (5,0) then continues unlimitedly,
if imagine this graph, it passed 1, III, and IV quadrants only, not II quadrant!
please visit the following website,
https://www.purplemath.com/modules/grphquad2.htm
try to understand if you cant write me private message, i will respond you as soon as I can.
if u have any questions, let me know!
