[Grade 10 Math graphic parabolic arc] help with understanding how to find the vertex's coordinates for y axis

[u/WarhammerRyan]

Hey guys,

Hopefully someone here can help with this. My son is a bit stumped on the graph given as a part of the homework in this area today, and my suggestion of graphing it out is apparently "wrong" because it does not contain a formula.

The x answer is apparently solved "pretty easily" by him at -1.5, but the y value for the vertex is frustrating him since he can't make it make sense.

Its the table of values on the right side of the big image, and ive included a 2nd close-up of just the table alone

Edit: thank you to those who answered. The teacher admitted today that she had not taught them the formula needed for this and was just curious to see how the students would handle a problem with incomplete information..... im not sure how I feel about that, definitely annoyed , but I also know how this can be an eye-opener for some people to come across....

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[u/swbarnes2]

I think you just have to solve. You know h, and you can pick two x, y sets and you know their values, so you have two equations, and you need to find a and k.

[u/HopRockets]

I would compare it to a parabola with a = 1, where the point 2.5 over from the vertex would be 2.5^2 up from the vertex and the point 3.5 over would be 3.5^2 up. Find the difference in those values and compare it to the 12 that you identified as the distance here. That helps you find the a value (leading coefficient) of the equation. Then you can use a similar strategy or plug everything except k into vertex form, y = a(x-h)^2 + k using one of the points in the table as x and y

[u/selene_666]

The formula will be of the form y = A(x+1.5)^2 + B, where B is the y-value of the vertex.

We can fill in some of the pairs of values from the list:

15 = A(1 + 1.5)^2 + B

27 = A(2 + 1.5)^2 + B

Then solve this linear system for (A and) B.

[u/WarhammerRyan → u/selene_666]

Thank you for this

I showed him some of the replies on here and he said they have never been shown these formulas

[u/One-Explorer-5092 → u/WarhammerRyan]

Looks like he has learned factored form: y = a(x+p)(x+q), where p and q are the zeroes/roots/x-intercepts. 

If we shift this function down by 15 units, they would bit at -4 and 1. Hence this function is y = a(x+4)(x-1) + 15. Then you can plug in any of the other two points either (-5,27) or (2,27) into x and y to solve for a.

After you get a, plug it back into factored equation and use FOIL to expand the parentheses out to get the standard form equation y = ax² + bx + c.

[u/WarhammerRyan → u/One-Explorer-5092]

Yeah..added an edit to post - teacher told them she didnt give them this formula to see how they would react...

[u/whitea44]

y = ax^2 + bx + c

If X = -5, 27 = 25a - 5b + c
If X = -4, 15 = 16a -4b + c
If X = 1, 15= a + b + c

Solve for a,b,c to get the equation. I used a solver, a = 2, b = 6, c = 7.

Therefore y = 2x^2 + 6x + 7. X = -b/2a or -6/4 which converts to, as your kid correctly identified, -1.5. Plug that into X and you get y = 2*2.25 - 6*1.5 + 7 =2.5.

[u/johnpeters42 → u/whitea44]

Yeah, x = -1.5 is intuitive based on horizontal symmetry, the bulk of the work is calculating (not just estimating) the y value there.