write an equation for a function that has a graph with the shape of y=x^2, but upsidedown and shifted right 6 units and up 4 units. ​

y - 4 = -(x - h)^2

Step-by-step explanation:

We begin with the stock function y = x^2.  Its graph is a parabola with vertex at (0, 0) and opening up.  The more general vertex form of the graph is

y - k = a(x - h)^2, where the point (h, k) is the vertex and 'a' determines the direction of opening as well as by how much the parabola is stretched vertically.

In this problem the vertex is (6, 4); that is, the vertex is shifted to the right by 6 units and upward from there by 4 units.  The coefficient 'a' is -1, indicating that the graph opens downward.

So we have the equation y - 4 = -(x - h)^2.  Without more information we cannot tell whether the graph is stretched vertically or not.

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."