1.which side of AFED does AB in AABC correspond to?
1 answer:
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There are a lot of steps involved in this, so pay attention. First step is to expand the squared quantity and FOIL it out, like this:
and
We are going to combine all the like terms now and get them all on one side of the equation:
Now we are going to complete the square on the polynomial in order to find the vertex. Do this by first setting the equation equal to 0 and then moving the constant over to the other side of the equals sign, like this:
and now factor out the negative sign (cuz negative signs are a pain):
. To complete the square, you take half the linear term, square it, and add it in to both sides. Our linear term is 6x. Half of 6 is 3, and 3 squared is 9. That's easy to add in on the left side, but we cannot forget that fact that we factored out a negative 1, and that the negative 1 is still there and has to be taken into consideration when we "add" in a 9 to the other side. We actually multiply the negative 1 times the 9 and that's what's added in to the right:
What you do when you complete the square is create a perfect square binomial on the left, which we have and which looks like this:
When we move the 3 back over to be with its mates (the 3 is the y coordinate for the vertex), we have the actual sign of the y coordinate. The number inside the parenthesis with the x is the x coordiante of the vertex in the form
. So the vertex of your problem is (-3, 3). The negative outside the parenthesis just indicates to us that the parabola is an upside down one, like a mountain instead of a valley.
We are going to combine all the like terms now and get them all on one side of the equation:
Now we are going to complete the square on the polynomial in order to find the vertex. Do this by first setting the equation equal to 0 and then moving the constant over to the other side of the equals sign, like this:
What you do when you complete the square is create a perfect square binomial on the left, which we have and which looks like this:
When we move the 3 back over to be with its mates (the 3 is the y coordinate for the vertex), we have the actual sign of the y coordinate. The number inside the parenthesis with the x is the x coordiante of the vertex in the form
Answer:
Step-by-step explanation:
Here's a step by step tutorial on your calculator on how to do this.
Hit "stat" then "1: Edit". If there are numbers there, arrow up to highlight L1, or L2, or wherever there are numbers. Hit "clear" then "enter" and the numbers will be gone. In L1, enter the Practice throws values. Press 3 then enter, then 12 then enter, then 6 then enter, etc. til all of them are in L1's column. Then arrow over to L2 and do the same with entering all the Free Throw values.
When you're done, hit "stat" again, then arrow over once to "calc" and #4 should say LinReg. That's a linear regression equation. If you have a TI 83, just hit enter and you'll get the equation in the form y = mx + b. If you have a TI 84 or 84+, you'll need to arrow down to the word "calculate" and then you'll see the equation.
One thing...if you have not turned your diagnostics on, you wont be able to see the coefficient of determination (the r-squared value). To make sure it's on:
Hit 2nd, then 0. You have opened up the catalog which lists every single thing your calculator can do in alphabetical order. The hit the button UNDER the MATH button (x to the negative 1) and scroll down until you see "diagnosticsON" and hit enter twice.
If you need to recall the linear regression equation, hit "stat", then "calc" then either enter or calculate and you'll get the linear equation again and an r value and an r-squared value. The closer that number is to 1, the better the data fits that model. I got that the equation is
y = 2.352x - .852 with an r-squared value of .844
To get the quadratic regression equation, hit "stat" then "calc" then choose #5, QuadReg. Repeat the process to calculate your equation. Mine was
And for the exponential regression, choose ExpReg (mine is under 0. Yours may be some other number. Just arrow down til you find it, it's there!) My ExpReg equation was
It appears that the r-squared value is the highest in the quadratic regression equation, so that's the best fit.