What is the answer /
1 answer:
You might be interested in
<u>Answer:</u>
f(x) = 7 is an even function.
<u>Step-by-step explanation:</u>
Any function is even when and gives a graph that is symmetric with respect to the y-axis.
We are given these options and we have to check whether which of them is an even function:
and - This function is not even
and - This function is not even
and - This function is not even
and - EVEN FUNCTION
We get the same value for and
, therefore f(x) = 7 is an even function.
Alright, if you can't remember what the quadratic formula is, I'll put it here for you:
(The plus and minus sign next to the -b are supposed to be under and over each other)
Now, all you have to do here is look at the question, and plug in the values into that tediously large equation. "But HotDogMan, what am I supposed to put in for the letters?!", Fret no more, as I will show you what the letters are.
'a' is the first number. You can find it next to the
in the equation. The 'b' is the next number over, which you can find next to the x. And finally, 'c' is the last number that is by itself to the right. Here is an example to show:
In this case, 'a' would be 6, 'b' would be -4, and 'c' would be 5. Now, if the order was different, something like this instead:
'a' would still be 6, 'b' would still be -4, and 'c' would still be 5. What matters the most here is what number is next to the x^2 , x , and what number is by itself. The final form looks like this:
Ok, I promise I'm almost done talking here. The second step now is just to remember that big equation up there and plug in 'a', 'b', and 'c'.
But before you do that, I wanna show you a cheat code real quick:
(Yes, cheat codes exist even in math)
(Notice how this snippet is actually part of the bigger equation at the top.)
Plug in the numbers from the question into this equation, and either 3 thee things will happen:
(1.)
That means there are two solutions to the question. Plug the answer you got into the
part and continue to solve the quadratic equation.
(2.)
That means that there is only one solution. At this point, you can use this equation to find that solution:
(3.)
If you get a negative number, you can stop right here. This means that there are no solutions to the question. At this point, you can write "No Solutions" on the question, and you should be good.
Now give me a couple of minutes, I will do the first one you have on the paper as an example for you there, and edit this when I finish:
So we have:
a = 5
b = -2
c = -7
Step 1.
We use the cheat code equation, (which is actually called the determinate btw)
This is equal to positive 144, so we plug this into the quadratic equation above.
Step 2.
The square root of 144 is 12, and 2×5 is 10, so we have something that looks like this:
Step 3.
[tex] \frac{2 +- 12}{10} [\tex]
The plus and minus means you need to split the answer into two different parts. The first one will be adding, the other will be subtracting. It will look something like this:
Step 4.
[tex] \frac{2+12}{10} \frac{2-12}{10} [\tex]
From there, you can simplify each fraction to get both solutions to the question.
I hope this helps, if you have any questions, let me know in the comments, I'll do my best to help and explain.
(The plus and minus sign next to the -b are supposed to be under and over each other)
Now, all you have to do here is look at the question, and plug in the values into that tediously large equation. "But HotDogMan, what am I supposed to put in for the letters?!", Fret no more, as I will show you what the letters are.
'a' is the first number. You can find it next to the
in the equation. The 'b' is the next number over, which you can find next to the x. And finally, 'c' is the last number that is by itself to the right. Here is an example to show:
In this case, 'a' would be 6, 'b' would be -4, and 'c' would be 5. Now, if the order was different, something like this instead:
'a' would still be 6, 'b' would still be -4, and 'c' would still be 5. What matters the most here is what number is next to the x^2 , x , and what number is by itself. The final form looks like this:
Ok, I promise I'm almost done talking here. The second step now is just to remember that big equation up there and plug in 'a', 'b', and 'c'.
But before you do that, I wanna show you a cheat code real quick:
(Yes, cheat codes exist even in math)
(Notice how this snippet is actually part of the bigger equation at the top.)
Plug in the numbers from the question into this equation, and either 3 thee things will happen:
(1.)
That means there are two solutions to the question. Plug the answer you got into the
part and continue to solve the quadratic equation.
(2.)
That means that there is only one solution. At this point, you can use this equation to find that solution:
(3.)
If you get a negative number, you can stop right here. This means that there are no solutions to the question. At this point, you can write "No Solutions" on the question, and you should be good.
Now give me a couple of minutes, I will do the first one you have on the paper as an example for you there, and edit this when I finish:
So we have:
a = 5
b = -2
c = -7
Step 1.
We use the cheat code equation, (which is actually called the determinate btw)
This is equal to positive 144, so we plug this into the quadratic equation above.
Step 2.
The square root of 144 is 12, and 2×5 is 10, so we have something that looks like this:
Step 3.
[tex] \frac{2 +- 12}{10} [\tex]
The plus and minus means you need to split the answer into two different parts. The first one will be adding, the other will be subtracting. It will look something like this:
Step 4.
[tex] \frac{2+12}{10} \frac{2-12}{10} [\tex]
From there, you can simplify each fraction to get both solutions to the question.
I hope this helps, if you have any questions, let me know in the comments, I'll do my best to help and explain.