Question

. Find a quadratic model for the set of values: (-2, -20), (0, -4), (4, -20). Show your work.

Answer 1

You need a system of 3 equations to solve for the 3 unknowns in our parabola, standard form of $y=ax^2+bx+c$.  Since the second point there has an x of 0, we'll start there to make it easy.  In this point, x = 0 and y = -4.  $-4=a(0)^2+b(0)+c$.  That gives us our first value...c = -4.  Let's do the first equation now the same way, but this time we have a c value to sub in: $-20=a(-2)^2+b(-2)+c$ and $-20=4a-2b-4$ which simplifies to $-16=4a-2b$.  We'll use that in a bit.  Let's do the third point now the same way. $-20=a(4)^2+b(4)+c$ and $-20=16a+4b-4$ which simplifies to $-16=16a+4b$.  Now we have a new system of equations.  We need to solve for a and b.  Let's multiply the 2nd equation by -4 to get rid of the a terms.  Doing that we have $64=-16a+8b$ and we will add that to $-16=16a+4b$.  The a terms cancel each other out leaving us with 48=12b and b = 4.  Now we'll sub that b into one of the equations in terms of a and b and solve for a.  -16 = 16a + 4(4) and -16=16a+16.  Subtracting 16 from both sides and we have -32=16a and a = -2.  Here's what we have for our values now:  a = -2 b = 4, c = -4.  So the quadratic in standard form is $y=-2x^2+4x-4$.  And you're done!