Thread: quadratic equations

hi all, i need some help on finding quadratic equations from points on graphs, could someone please please explain how i would determine which general equation to use to solve.

1.) A parabola has a y intercept of 3 and x intercepts of 1 and 3, find the rule dictating the parabola.

a.) someone please explain how to pick a formula to use

2.) a parabola passes through points (-3,9) find the rule dictating the parabola.

a.) someone please explain how to pick a formula to use

also at the start of the exercise there is an example like this: a parabola passes through points (0,3) and (-3,1) it says this equation is in the from y =ax^2 + c... What the hell? i nvr learnt any ax^2 +c

pls help i have a test this coming thursday

Originally Posted by andrew2322

hi all, i need some help on finding quadratic equations from points on graphs, could someone please please explain how i would determine which general equation to use to solve.

1.) A parabola has a y intercept of 3 and x intercepts of 1 and 3, find the rule dictating the parabola.

Mr F says: Use the model . B = 1, C = 3 (why?). To get A substitute the given y-intercept, that is, the point (0, 3).

a.) someone please explain how to pick a formula to use

2.) a parabola passes through points (-3,9) find the rule dictating the parabola. Mr F says: Insufficient data to get a unique rule.

a.) someone please explain how to pick a formula to use

also at the start of the exercise there is an example like this: a parabola passes through points (0,3) and (-3,1) it says this equation is in the from y =ax^2 + c... What the hell? i nvr learnt any ax^2 +c

Mr F says: They're giving you the heads-up that b = 0 in the standard form! Run with the ball!! Substitute the two given points to get two equations in a and c. Solve these two equations simultaneously.

pls help i have a test this coming thursday

..

huh? im like so lost... how can y = A(x-c)(x-b) work?gah pls go over it again with the actual answers pls? nd how to select which formula to use pls

Originally Posted by andrew2322

huh? im like so lost... how can y = A(x-c)(x-b) work?gah pls go over it again with the actual answers pls? nd how to select which formula to use pls

Consider ....... y = A(x - B)(x - C) has x-intercepts found from the equation

A(x - B)(x - C) = 0

=> x - B = 0 or x - C = 0

=> x = B or x = C.

So if you know the x-intercepts, you use the model y = A(x - B)(x - C) because knowing the x-intercepts means you know the value of B and C .....

So for the question:

Originally Posted by andrew2322

1.) A parabola has a y intercept of 3 and x intercepts of 1 and 3, find the rule dictating the parabola.

you would use this model and you know that B = 1 and C = 3.

So y = A(x - 1)(x - 3).

Now substitute the y-inetercept, that is, the point (0, 3), that is, x = 0 and y = 3:

3 = A(0 - 1)(0 - 3) = A(-1)(-3) = 3A => A = 1.

So y = (x - 1)(x - 3).

wow that just hit me in the face, how cud i have not seen the obvious... wow ur really smart. and one last thing, cud u pls explain the values B and C and how they are different from X? and also, there is another example as y = ax^2? how did that come about? the question is the parabola passes through points (2,5) find the equation that dictates the graph.

Originally Posted by andrew2322

wow that just hit me in the face, how cud i have not seen the obvious... wow ur really smart. and one last thing, cud u pls explain the values B and C and how they are different from X?

They are the x-intercepts. In other words, the value of x where the parabola cuts the x-axis. In other words, when y = 0, x = B or x = C.

Try working backwards from the answer y = (x - 1)(x - 3)to see how it all fits together.

there is another example as y = ax^2? how did that come about? the question is the parabola passes through points (2,5) find the equation that dictates the graph.

Originally Posted by andrew2322

there is another example as y = ax^2? how did that come about? the question is the parabola passes through points (2,5) find the equation that dictates the graph.

They give you the model - on a platter. Don't question your good luck, just use it!!!!

Substitute the given point, that is x = 2 and y = 5:

5 = a (2)^2 = 4a => a = 5/4.

Therefore y = (5/4) x^2.

could you please tell me how, if i didnt have good luck, know what equation to use?

Originally Posted by andrew2322

could you please tell me how, if i didnt have good luck, know what equation to use?

You wouldn't. Without being given the model, a single point is insufficient data to get a unique equation. That's why you were given the equation (so it wasn't really good luck .... it was actually absolutely essential).

ok just one FINAL question

Y intercept: (0,5)
X intercept: (4,0)

Find the equation, please give me step by step working out with the final answer, thanks

Originally Posted by andrew2322

ok just one FINAL question

Y intercept: (0,5)
X intercept: (4,0)

Find the equation, please give me step by step working out with the final answer, thanks

Insufficient data to find a unique equation. Is the x-intercept also a turning point?