At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
so let x = 4 be a zero then x -4 is a factor so you just need another factor... then multiply them together to get the standard form or the quadratic
for the 2nd function if x = -4 is an intercept or zero, then x + 4 is a factor so same as above... or you could combine both factors.
how do I get the other factor @campbell_st
well you can use the same value... the quadratic is a perfect square so if you want another factor... let an x intercept by (-1, 0) so what factor would you get from this point..?
well a zero or intercept is at x = -1 so then x + 1 is a factor you can use this for both questions y = (x-4)(x+1) part 2 y = (x +4)(x + 1)
oh I see
I got the answer wrong @campbell_st
it asks for a parabola that opens upward and one that opens downward
well you can have a perfect square \[y = (x -4)^2\] is there any other information in the question that you didn't include..?
the example of \[y=(x -4)(x+1) = x^2 -3x - 4\] will open upwards, because the coefficient of x^2 is positive.
yeah I put that and it says its wrong
Idk why because the x intercept is 4
Find two quadratic functions, one that opens upward and one that opens downward, whose graphs have the given x-intercepts. (There are many correct answers.) (−4, 0), (4, 0)
ok... so the intercepts are x = -4 so x + 4 is a factor x = 4 so x - 4 is a factor so the quadratic is \[y = a(x -4)(x +4)\] or \[y = a(x^2 - 16)\] for upwards make a a positive number and multiply, 1 is probably easiest for downwards pick a negative number for a so maybe a = -1 and then multiply out... that should help you get the correct answer
ok let me plug it in