Can someone please help me! Math methods units 1&2

[Rose34]

Families of quadratic polynomial function:

It is a graph that has the point (-1,5) and (-3,0) and a turning point of (-1,5), and we are asked to find the equation of the parabola?

I tried to use the equation of y=ax^2+c and sub the points in it to create a simultaneous equations:

First I subbed (-3,0):
0=a(-3)^2+c
c=-9a (first equation)

Then I subbed (1,0):
0=a(1)^2+c
c=-a (second equation)

Then I subbed eq1 and eq2 together
-9a=-a
add a to both sides which becomes -8a=0  divide -8 for both sides thus a=0

Then to find c I subbed into y=ax^2+c and using the Turning point (-1,5) i subbed it as x and y

5=-8(-1)^2+c
c=13

lastly I put everything together into y=ax^2+c
y=-8x^2+13
When I checked the answer in the book it is apparently y=-5/4x^2-5/2x+15/4 

As you can see my answer is way off so i do not know where i went wrong and this topic is completely new to me so i have no idea what i did wrong and if i subbed into the correct equation of y=ax^2+c or I have to use another formula?

If you know what to do please help me..
Thanks in advance

[yesh.weerakkody]

Heyoo! I’ll try to help but of course I may be wrong...

I think what went wrong was with finding the dilation part since it cannot be 0(the graph cannot exist at 0).

This is the working out I did:
Sub (-1,5) into the turning point equation since that is the easiest way to find all the variables (because the t.p. is already given...
y= a(x-(-1))^2 + 5... which would simplify into,
y= a(x+1)^2 + 5
Then you can sub (-3,0) into the equation to find a.
0=a(-3+1)^2 + 5
  = a(-2)^2 +5
  = 4a + 5
-5=4a
-5/4=a

Then you can sub in to find rest.
y=-5/4(x+1)^2 + 5
  =-5/4(x^2+2x+1) +5
  =-5/4x^2 -10/4x -5/4 + 20/4
  =-5/4x^2 - 5/2x +15/4
Hope this helps!

[BiggestVCESweat]

y=a x^2 + c should never be used except in some very rare cases.
The correct formulas are:
y = ax^2 + bx + c (should be used when three random points are known; works with any points but is slow)
y = a(x - h)^2 + k (used when turning point is known; turning point is (h,k))
y = a(x-d)(x-e) (should be used when x-intercepts are known; d and e are the x-intercepts.)

In this case although all the formulas can be used in this case, I would definitely use the third as in this scenario it would be the easiest to use. There are however scenarios where the others would be easier to use.

[Rose34]

Heyoo! I’ll try to help but of course I may be wrong...

I think what went wrong was with finding the dilation part since it cannot be 0(the graph cannot exist at 0).

This is the working out I did:
Sub (-1,5) into the turning point equation since that is the easiest way to find all the variables (because the t.p. is already given...
y= a(x-(-1))^2 + 5... which would simplify into,
y= a(x+1)^2 + 5
Then you can sub (-3,0) into the equation to find a.
0=a(-3+1)^2 + 5
  = a(-2)^2 +5
  = 4a + 5
-5=4a
-5/4=a

Then you can sub in to find rest.
y=-5/4(x+1)^2 + 5
  =-5/4(x^2+2x+1) +5
  =-5/4x^2 -10/4x -5/4 + 20/4
  =-5/4x^2 - 5/2x +15/4
Hope this helps!

Thank you so much that helped me a lot!!

[Rose34]

y=a x^2 + c should never be used except in some very rare cases.
The correct formulas are:
y = ax^2 + bx + c (should be used when three random points are known; works with any points but is slow)
y = a(x - h)^2 + k (used when turning point is known; turning point is (h,k))
y = a(x-d)(x-e) (should be used when x-intercepts are known; d and e are the x-intercepts.)

In this case although all the formulas can be used in this case, I would definitely use the third as in this scenario it would be the easiest to use. There are however scenarios where the others would be easier to use.
[/quote

Thank you so much!! Do you have a way to remember these formulas?

[Rose34]

Heyoo! I’ll try to help but of course I may be wrong...

I think what went wrong was with finding the dilation part since it cannot be 0(the graph cannot exist at 0).

This is the working out I did:
Sub (-1,5) into the turning point equation since that is the easiest way to find all the variables (because the t.p. is already given...
y= a(x-(-1))^2 + 5... which would simplify into,
y= a(x+1)^2 + 5
Then you can sub (-3,0) into the equation to find a.
0=a(-3+1)^2 + 5
  = a(-2)^2 +5
  = 4a + 5
-5=4a
-5/4=a

Then you can sub in to find rest.
y=-5/4(x+1)^2 + 5
  =-5/4(x^2+2x+1) +5
  =-5/4x^2 -10/4x -5/4 + 20/4
  =-5/4x^2 - 5/2x +15/4
Hope this helps!

Sorry I have a question, you are absolutely correct, I just want to know why do we need to expand in the last step, y=-5/4(x+1)^2 + 5, like can we just leave it without expanding?

[colline]

Sorry I have a question, you are absolutely correct, I just want to know why do we need to expand in the last step, y=-5/4(x+1)^2 + 5, like can we just leave it without expanding?

Generally when finding the equation of a parabola we'd (by convention) leave it in the form y=ax²+bx+c

Thank you so much!! Do you have a way to remember these formulas?

The best and most foolproof way is to just do HEAPS of relevant questions continuously for a couple of days, then they'll be permanently carved into your memory.

[Rose34]

Generally when finding the equation of a parabola we'd (by convention) leave it in the form y=ax²+bx+c
The best and most foolproof way is to just do HEAPS of relevant questions continuously for a couple of days, then they'll be permanently carved into your memory.

Thank you so much!!
Do you know where can I find resources for this kind of question?

[colline]

Thank you so much!!
Do you know where can I find resources for this kind of question?

No worries

Your teacher is generally the first person to ask - if he/she doesn't have any, your school's VCE methods coordinator / head of VCE maths (whatever your school calls them) should have plenty.

Otherwise, there would be many quadratics worksheets online as well, though they most likely won't be VCE specific. Good luck!

[Rose34]

No worries

Your teacher is generally the first person to ask - if he/she doesn't have any, your school's VCE methods coordinator / head of VCE maths (whatever your school calls them) should have plenty.

Otherwise, there would be many quadratics worksheets online as well, though they most likely won't be VCE specific. Good luck!

Thank you