Solution to 1 & 16 and request for 13

For number 1, we can take the given equation and plug in the point, say, (-2,8), and use this to solve the equation for k in terms of m...

y = x^2 + kx + m , such that (x,y) = (-2,8)

8 = (-2)^2 + k(-2) + m

k = -2 + 1/2*m

Then plug in the solution for k to the original equation, this time using (4,-6) as x and y, in order to solve for m...

y = x^2 + kx + m , such that (x,y) = (4,-6)

-6 = 4^2 + (-2 + 1/2*m)*4 + m

m = -14/3

Now plug in the solution for m into the k equation, and finally, find k + m...

k = -2 + 1/2* (-14/3)

k = -13/3

k + m = -13/3 + -14/3 = -27/3

= -9

For number 16, we can take the given equation and assume that using the point P (1,2), there is a point on the graph y = 8 - x^2 that is (x,2), and plug the y value into the equation...

y = 8 - x^2

2 = 8 - x^2

x = +/- sqrt(6)

Then solve for the 2 cases when x = sqrt(6) and when x = -sqrt(6), and we get y values of 2 and 14, respectively, giving us (sqrt(6),2) and (-sqrt(6),14) for points on the graph.

I was also wondering if someone had a sol'n for number 13. I always have problems w/ these.