Question

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

Answer 1

are you supposed to graph them?

Answer 2

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.

Answer 3

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

Answer 4

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..

Answer 5

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

Answer 6

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!

Answer 7

A one line Mathematica solution is attached.

Answer 8

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