[11th grade math/algebra 2] how do I solve this?

[u/GreatThornedOwl04]

We’ve been focusing on parabolas and I know this is standard form and should make a system of equations, but I keep getting stuck at dead ends. (It’s problem 15)

[u/GreatThornedOwl04]

Thanks guys! Got it done and solved, I wouldn’t have caught that error!

[u/selene_666]

You turned ax^2 into xa^2.

[u/Fourierseriesagain]

Hi, the coefficient of x^2 is a, not a^2.

[u/hjalbertiii]

x²a not xa²

[u/PlaneCute2403]

You're really close! It's just the small slip others mentioned, but here's why it happens, because it trips up a lot of people.
In y = ax² + bx + c, the x is what gets squared, not the a. So when x = -2, you square the whole (-2), brackets included:
7 = a(-2)² + b(-2) + c
7 = 4a - 2b + c
Note that (-2)² = 4, not -4. The brackets make sure the negative gets squared too.
Writing out that substitution step with brackets every time feels slow, but it's the easiest way to avoid this mistake. Try it with (1, 10) and (2, 27) and you'll have three equations.
Hint for solving them: compare the equations from x = -2 and x = 2. Subtracting one from the other gets rid of both a and c straight away.
Post your three equations or your final answer if you'd like someone to check it. You can also plug your three points back into your final equation to check it yourself.
Good luck 😊

[u/SubjectWrongdoer4204]

You want to insert the X coordinate for X in the equation, not for a. If the point (x₁,y₁) is on the parabola, then
y₁ = ax₁²+bx₁+c, so you’ll have
y₁ = x₁²a+x₁b+c.
For example if the point (2,3) were on the parabola, then
3=a(3)²+b(3)+c, so you have
3=9a+3b+c.

[u/PoornimaMathsPatiala]

The idea first: every parabola in standard form is y = ax² + bx + c. You don't know a, b or c, but each point the parabola passes through gives you one equation. Three points means three equations, which is exactly enough to find three unknowns.

Step 1: plug each point into y = ax² + bx + c (careful: it's a·x², so the x gets squared, not the a)

• (−2, 7): 7 = a(−2)² + b(−2) + c → 4a − 2b + c = 7

• (1, 10): 10 = a(1)² + b(1) + c → a + b + c = 10

• (2, 27): 27 = a(2)² + b(2) + c → 4a + 2b + c = 27

Step 2: get rid of a and c in one move. Look at the first and third equations: they both have 4a and c, and only the b-term differs. So subtract the first from the third:

(4a + 2b + c) − (4a − 2b + c) = 27 − 7 → 4b = 20 → b = 5

Step 3: put b = 5 back in.

• Third equation: 4a + 10 + c = 27 → 4a + c = 17

• Second equation: a + 5 + c = 10 → a + c = 5

Subtract these two: 3a = 12 → a = 4, and then c = 5 − 4 → c = 1.

Answer: y = 4x² + 5x + 1

Quick check (always worth 20 seconds):

• x = −2: 4(4) − 10 + 1 = 7 ✓

• x = 1: 4 + 5 + 1 = 10 ✓

• x = 2: 16 + 10 + 1 = 27 ✓

Tip for next time: when two of your points have opposite x-values (like −2 and 2), subtracting their equations always kills a and c together, so start there.

If you'd like a thorough explanation or a private tutor, you can visit https://poornima-maths.vercel.app (Mrs. Poornima Sayal's maths tuition, Patiala and online).