Question 105257
One way to do these is to substitute the x- and y-values from the given points into the equation, thus giving you a system of equations with two unknowns (a and k). Then solve the system of equations for the two unknowns.
{{{y = a(x+1)^2+k}}} Substitute x = 1 and y = 11 from the first point (1, 11) to get the first equation.
{{{11 = a(1+1)^2+k}}} Simplify.
{{{11 = 4a+k}}} 
Second equation:
{{{y = a(x+1)^2+k}}} Substitute the x- and y-values from the second point (2, -19) to get the second equation:
{{{-19 = a(2+1)^2+k}}} Simplify.
{{{-19 = 9a+k}}} 
So now you have the two equatios:
1) {{{11 = 4a+k}}} and
2) {{{-19 = 9a+k}}} You can solve these by any of the accepted methods for solving systems of equations. Let's use elimination by subtracting equation 1) from equation 2) and solving for a.
{{{-30 = 5a}}} Divide both sides by 5.
{{{a = -6}}} Now substitute this value of a into either equation 1) or equation 2) and solve for k. Let's use equation 1)
{{{11 = 4a+k}}} Substitute a = -6
{{{11 = 4(-6)+k}}}
{{{11 = -24+k}}} Add 24 to both sides.
{{{k = 35}}}
Check: Into the original equation, substitute a = -6 and k = 35.
{{{y = -6(x+1)^2+35}}}
Now let's see if the two given points (1, 11) and (2, -19) satisfy this equation.
{{{11 = -6(1+1)^2+35}}}
{{{11 = -6(4)+35}}}
{{{11 = -24+35}}}
{{{11 = 11}}} Ok for the first point.
{{{-19 = -6(2+1)^2+35}}}
{{{-19 = -6(9)+35}}}
{{{-19 = -54+35}}}
{{{-19 = -19}}} Ok for the second point.