Given the following points on a quadratic function; (-3, 17), (-1,-3), (4,52). Show that (1.1)is also on that graph.

are you supposed to graph them?

My approach would have been to create a system of equations from the general form \(y = ax^2 + bx + c\) and plug in each point for three 3-variable equations, and then solve the system for the a, b, and c values. Then, if you plug in (1,1), it should be true as well with those (a,b,c) values.

So, if we plug in the point values and create this system, we get: \[ y = ax^2 + bx + c \qquad (\neg 3, \ 17), \ (-1, \ \neg3), \ (4, \ 52) \\ \quad (\neg 3, \ 17) \implies a(-3)^2 + b(-3) + c = 17 \\ \quad (\neg 1, \ \neg 3) \implies a(-1)^2 + b(-1) + c = \neg 3 \\ \quad (4, \ 52) \implies a(4)^2 + b(4) + c = 52 \\ \\ \text{So, we simplify and the system becomes:}\\ \begin{align} \quad 9 & a - 3 b + c = 17 \\ \quad & a - \ \: b + c = \neg 3 \\ \quad 16 & a + 4 b + c = 52 \end{align}\]

Then, we can pretty much reduce it into a two-variable system by elimination. Multiply (a - b + c = -3) by -1 and then add this to both the first and third equations. It should just become perfect for another elimination from there, and you just have to do a bit of substitutions for the actual solution..

a - b + c = -3 Multiply by -1 -a + b - c = 3 Add to (1) -a + b - c = 3 9a - 3b + c = 17 ----------------- 8a - 2b = 20 Then, divide off 2 from both sides 4a - b = 10 Add to (3) -a + b - c = 3 16a + 4b + c = 52 ----------------- 15a + 5b = 55 Then, divide off 5 from both sides 3a + b = 11 You can then add these new equations together to eliminate b: 3a + b = 11 4a - b = 10 ------------- 7a = 21 Then, divide off 7 a = 3 For the remaining two variables, it's just substitution from here. 3(3) + b = 11 9 + b = 11 b = 2 -3 + 2 - c = 3 -1 - c = 3 -c = 4 c = -4 (a,b,c) = (3,2,-4) --> Put back into original general form

y = ax^2 + bx + c y = 3x^2 + 2x - 4 (1,1) should work here! 1 = 3(1) + 2(1) - 4 1 = 3 + 2 - 4 1 = 5 - 4 1 = 1 Yes!

A one line Mathematica solution is attached.