Math algebra 2 polynomial functions

Question

Find a polynomial function whose graph passes through (-1,-2),(0,-4),(1,-2),(2,16)

Answer 1

a piece meal function works, which is mostly quad

y=2x^2 -4 for x<1

y = 18x -16 for 1<x<2

a line through (2,16) and (1,-2) has slope = 18

y+2 = 18(x-1)= 18x-18

y = 18x -16

but for a lower degree "polynomial" piece mean function

y=16, -2 and -4 works. zero degree polynomial

or run line segments through the 1st 2 pionts, 2n secod pair and 3rd pair

for a linear polynomial piece meal function. y=18x-16 is the 3rd pair's line segment equation

y=2x +4 for the 2nd, y =-2x +4

f(x) = {-2x+4, for x<-0

{2x +4, for 0< x < 4

{18x-16 for x> 4

or for not piece meal

lowest degree graph would be

y = 22(x^2-2)^2 -24

read further,

4th degree polynomials can be thought of as degenerate quadratics

looks like an upward opening parabola quadratic, with vertex = (0,-4) = minimum point

with axis of symmetry x=0

y = a(x-0)^2 - 4 in vertex form, y=a(x-h)^2 +k with (h,k)=vertex = (0,-4)

plug in one of the points, (2,16), x=2, y=16, to solve for a, the coefficient of the leading term, ax^2.

16 = a(2^2) -4

16= 4a -4

4a = 20

a =20/4 = 5

y = 5x^2 -4

try one of the other points, (1,-2), plug it in

-2= 5(1^2) -4= 5-4 = 1

-2 doesn't = 1

that didn't work

but not that far off, relatively

odds are the original problem was or should have been

with the points (-1,1), (0,-4), (1,1) and (2,16)

or as (-1,-2), (0,-4), (1,-2) and (2,4)

4 instead of 16 seems likely, given a quadatic as 4^2 = 16

but a 2 showing up when 1 was intended, can happen as they're adjoining on a key board.

it's definitely not a linear relation,

not cubic or an odd degree, as the points don't have that likely shape

possibly 4th degree, or a higher even degree, but for an algebra 2 class that seems very unlikely for higher degrees

plot the points. they look too much like a parabola, just off a little, to have been intended as other than a quadratic

quadratic is the usual type of relation in algebra 2 courses

actually there are an infinite number of equations and curves that could go through those 4 points, but none of them are a standard form that you would encounter in an algebra 2 class

A old standby is a sine wave function With a small enough period, and large enough amplitude, you can get the graph to pass through any set of points. y = AsinB(x+C) + D. Make A and B large enough, (x,y) will include any points, 4 or more, up to +infinity points. B varies inversely with the period.

Answer 2

By partial differnces the function is cubic.

f(x) = ax³ + bx² + cx + d

-4 = 0a + 0b + 0c + d

-2 = -a + b - d + d

-2 = a + b + c + d

16 = 8a + 4b + 2c + d

Can you solve the system for a, b, c, d, and answer?