Using TI-83 to get the Parabola?

Question

The question is:

Write the equation of the parabola shown "below" in f(x)=a(x-h)^2+k and in the form f(x)=ax^2+bx+c.

You may assume the leading coefficient is 1.

Now "below" the paper it shows the TI-83 graph screen "window" set to:

Xmin= -8
Xmax= 8
Xscl=1
Ymin= -8
Ymax= 8
Yscl= 1
Xres= 1

How am I supposs to come up with two equations...with so little information? Please help!!!

Answer 1

You need to do two things:
1. set your window
2. program the equations.

1. At the top there's a 'Window' button. You can set window parameters there. However, I'd recommend using the 'Zoon' button and chosing option 6: ZStandard. This will give you a good window. You can change it later.

2. The top left button says 'Y='. Press this and program equation 1 then 'enter'. This put you down one line so that you can program equation 2.

To run press the upper right hand button 'Graph'.

Also, go to the Texas Instruments website for help and to download an users manual.

Answer 2

well since you can assume that the leading coefficients is 1 meaning that the number 1 will be used for the letters a,b and c in your equation ax^2+bx+c because those letters represent coefficients in the formula. anywho, this is what it would look like as you would substitute 1 into the formula:

(1)x^2 + (1)x + (1) which will equal: x^2+x+1

then graph the new found equation into your calculator. Notice that when you graph it, the parabola does not seem to be in the center of the screen. What you have to do is find the vertex or the point in the middle of the parabola. you can do this by doin 2nd and calc on your TI-83 calculator and using the minimum option which is number 3. then create the bounds of where you want the calculator to calculate where the lowest point is or in your case the point in the middle of the parabola(vertex). First hit enter after aiming the cursor to the left side of the parabola close to the center then do the same on the right. after you do this the calculator will say Guess? so just ignore that and press enter again. The calculator will now give you the x and y coordinates of that point. Since you want to put you final answer in the

f(x)=a(x-h)^2+k form, then you now know how many units the parabola moved to the left. Use answer from calculator (x= -.5) and plug that into the formula. a(x--.5)^2+k then plug in the value of 1 for "a" and "k" since the problem told you so and fix the signs in the parenthesis. the answer sould look something like this:

(1)(x+.5)^2+(1) ==> then goes to: f(x)=(x+5)^2+1

I hope that helps you out.

Answer 3

here are some things to note without just giving you the answer:
the two equations are the same - the 2nd (ax^2+bx+c) is the same as the first equation if it is expanded/multiplied out.

as for getting the equation - everything you need to know to fill in the coefficients in equation one (a, h, k) you find by studying the parabola - primarily the location of its vertex. you should have equations in your textbook or a quick google-search can likely give them to you as well.

Answer 4

f(x) = ( x - h)^2 + k
y = x^2 - 2hx + h^2 .... a=1 b=2h c= h^2

y = x^2 + 3x + 2
&
y = x^2 -3x - 40