There seems to be a flaw with this question because it says that there are five x-intercepts but the given information only gives you 4 x-intercepts to work with.
Even means the graph is symmetric about the y-axis
The best answer is <span>A.(–6, 0), (–2, 0), and (0, 0)
because you do not have to worry about another point (0,0). Plus we need (-6,0) for it to be symmetric with (6,0).
Consider function f(x) = x²(x-6)(x+6)(x+2)</span>²(x-2)<span>². It is even and fits these conditions as it has x-intercepts at (6,0), (-6,0), (-2,0), (2,0), and (0,0). again, the question does not tell us the fifth x-intercept, so we need to assume that there is another one that needs to be there...and so (-2,0) must have (2,0) for it to be even as well.</span>
The discriminant tells you where the graph of the parabola goes through the x-axis, if at all. If the discriminant is negative there are no real zeros and the parabola does not cross or touch the x-axis; if the discriminant is positive the parabola will go through the x axis in 2 places; if the discriminant is 0 the parabola will touch the x-axis in 1 place. Our discriminant is 0 since the parabola only touches the x-axis but does not go through.
We use formula of BODMAS we starting by removing blackets by multiply (2)(3)and w get 6 and then by 2 we get 12 and add by 5 we get 17 and cross 17 to -36 and 17 change to -17 and then take -17-36=-53..
