if-hx3x-42-find-all-values-of-x-for-which-hx-12

Name: if hx3x 42 find all values of x for which hx 12
Uploaded: 2020-06-12
Duration: 1 min 11 s
Channel: Numerade
Description: If $h(x)=3(x-4)^{2},$ find all values of $x$ for which $h(x)=-12$

Question

If $h(x)=3(x-4)^{2},$ find all values of $x$ for which $h(x)=-12$

Nez Nikoo · Accepted Answer

Okay. Use equation. They&#x27;re telling us to find each of exit H of X is minus slow. So I put a minus 12 year and you see right away I can cancel this three this file submit. This becomes four. So now the equation is X minus four. Swear pickle. Two months. Yes. Okay, so now get a square root of each side. Get X minus four is speak well to plus minus and minus one becomes I four becomes 22 becomes too. I right and X is equal to four plus minus two I with many of these very straightforward. But when you have to know is if I have minus four, I really have a minus one and it four. And if you get a square root of this, what happens? You have I It&#x27;s a sign and four becomes too. So this is saying is to I think a little extra explanation again, but check. Thank you.

Ankit Gupta · Answer

According to the ocean we have given equation is at Joe X Physical dough four x minus X squared In discussion, we have also given Etch off A is equal to minors Carted toe, we find a first awful. We find out at Joe A the physical toe four a minus is well And then we put the value off edge off a physical to minus 32. When we put here minor started to is equal to four a minus is scwill. When we really in this question, there is four. A minus is where plus 32 we take minus common. In this equation we have a square minus four a minus 32 on we expand This a question. When we expand, we have isso will minus a day. Plus for a miner&#x27;s party toe is equal to zero. We take common a in discussion. We have a minus eight and with a common for in this side we have a minus eight is equal to zero. The factors comes out there is a plus four and the second factories a minus eight is equal to zero. We know that. According toa these caution we used the principle of zeros. Grab. There is a plus for is a little Vero and a minus. It is equal to zero. We find out the value of a is minus four and value of is eight. We can say that the value of a is minus four and it

Amy Jiang · Answer

Okay. We want to know one h of where did he put it? Falling is greater than six. Start by subtracting one on both sides to isolate our absolute value and how we can use our definition of absolute value. But it&#x27;s up into two terms. Okay? And I would solve for X so adding three and then dividing by two. We get this. Poor X is less than negative one. Okay, so this is our solution.

Peggy Radtke · Answer

functions tell us that for every value of X, that&#x27;s her input. And when we put that value into our equation, we&#x27;re going to get a output one input one output and functions could be written f of acts. We&#x27;ve also got functions that could be g of X. We could use a function h of acts, and all that just tells us is that we have different functions, and again, each of those functions have their corresponding equation, and it is going to be slightly different now for our functions. We have one input and we get one output. An example of this is if we had an equation such as F of axe equals two X squared minus four. Now we could find lots of different values of X. We could find f of zero. And what that&#x27;s telling us is that zero is our input, and we have to figure out what that output is. So if we were to replace all of our axes with zero and again that&#x27;s squared, that&#x27;s going to give us some kind of output. So here again is our input, and we&#x27;re solving for our output. Zero square 20 to 10 00 anything. So we have zero minus for and so our output is negative for so again, we have an input and an output. If we had f of one, we would replace acts with one. So it looked like this two times one squared, minus four. And again, we could solve that. So again, what&#x27;s in parentheses is our input value. And we replace that into our function, which is our equation, and we get our output.

Amy Jiang · Answer

Okay, so we went to determine when are following function. H of X, which is able to falling, is good or than 18. So adding six humble sides. 18 06 We got 24 and now I&#x27;m using the definition of action volume. We get the falling now it&#x27;s all for X, so we&#x27;ll add four on both sides. So this is 28 and then we have to act is less than negative 20. Dividing by two on both sides. We get that X must be greater than 14. Four X must be less than negative 10. Okay, so we have the falling solutions.

Brandon Fox · Answer

Let&#x27;s substitute negative to infer h of X. So negative two equals negative, 1/2 ax minus six and let&#x27;s solve for the variable X. We&#x27;ll start by adding six on both sides, which we&#x27;ll get left with four equals a negative 1/2 acts and then divide both sides by negative one house. This is gonna leave us with a negative eight, so x equals negative eight.

Problem

Find the distance between each pair of points. If…

Problem 62 Easy Difficulty

If $h(x)=3(x-4)^{2},$ find all values of $x$ for which $h(x)=-12$

Answer

Related Courses

Introductory & Intermediate Algebra for College Students

Related Topics

Discussion

Top Algebra Educators

Grace H.

Heather Z.

Kristen K.

Michael J.

Algebra Courses

Absolute Value - Example 1

Absolute Value - Example 2

Recommended Videos

Watch More Solved Questions in Chapter 11

Video Transcript

Robert Blitzer

Introductory & Intermediate Algebra for College Students

Grace H.

Heather Z.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Introductory & Intermediate Algebra for College Students

Grace H.

Heather Z.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

Robert BlitzerIntroductory & Intermediate Algebra for College Students

Grace H.

Heather Z.

Kristen K.

Michael J.

Robert Blitzer

Introductory & Intermediate Algebra for College Students