Can someone please help me with a question?

- Liv1234

Can someone please help me with a question?

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- Liv1234

##### 1 Attachment

- Michele_Laino

the new function g(x) has to be equal to the old function f at point x+1

- Liv1234

What does that mean?

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## More answers

- Michele_Laino

in formula, my statements means:
g(x)=f(x+1)

- Liv1234

So kind of like for example y= mx+b?

- Michele_Laino

in that case, we can write this:
g(x9=m(x+1)+b

- Michele_Laino

oops..
g(x)=m(x+1)+b

- Liv1234

By the way, sorry if I seem slow with this, I'm just really bad with math.

- Michele_Laino

no worries! :)

- Liv1234

So, g(x)=m(x+1)+b

- Michele_Laino

yes!

- Liv1234

How would I solve the answer with that?

- Michele_Laino

I think you have to do the multiplication, like below:
g(x)=mx+m+b

- Liv1234

How do I do the multiplication with this equation?

- Michele_Laino

we can do the multiplication above using the distributive property of multiplication over addition

- Liv1234

I think I remember how to do that, is there a formula?

- Liv1234

So, (sorry to ask again) what numbers would be put in?

- Michele_Laino

the general formula, is:
\[\left( {a + b} \right)c = ac + bc,\quad \forall a,b,c \in \mathbb{R}\]

- Michele_Laino

you have to know the values of the coefficients m and b

- Liv1234

I feel really stupid to keep asking for help, but how would I find the coefficients?

- Michele_Laino

you can retrieve the value of those coefficients, from the text of your exercise

- Liv1234

The only number I had from the exercise is "1"

- Michele_Laino

in that case your exercise is not a numeric exercise

- Liv1234

Hang on, let me repost the question

- Michele_Laino

ok!

- Liv1234

##### 1 Attachment

- Liv1234

Okay, that's the question

- Michele_Laino

your answer, is:
\[g\left( x \right) = m\left( {x + 1} \right) + b = mx + m + b\]
and we have completed your exercise

- Liv1234

Wait, so what option would I choose? I'm confused.

- Michele_Laino

the first option

- Liv1234

Can you help me with another question?

- Michele_Laino

ok!

- Liv1234

##### 1 Attachment

- Michele_Laino

when we make a reflection with respect to the x-axis, then we can write the new function g(x) as below:
\[g\left( x \right) = - f\left( x \right)\]

- Liv1234

So, how would I solve?

- Michele_Laino

it is simple, you have to replace f(x) with its formula:
\[g\left( x \right) = - f\left( x \right) = - \left( {{x^2} + 5} \right) = ...?\]

- Liv1234

Would the answer be g(x)= -x^2+5?

- Michele_Laino

not exactly, since when you cancel the parentheses, you get this:
\[g\left( x \right) = - f\left( x \right) = - \left( {{x^2} + 5} \right) = - {x^2} - 5\]

- Liv1234

I'm confused D:

- Michele_Laino

every term inside the parentheses change its sign

- Michele_Laino

so we can write this:
\[ - \left( {{x^2} + 5} \right) = - {x^2} - 5\]

- Liv1234

So, the first option?

- Michele_Laino

yes!

- Liv1234

Can you help me with another one?

- Michele_Laino

ok!

- Liv1234

##### 1 Attachment

- Michele_Laino

here we have to consider all points whose y-coordinate is less the y coordinate of the points which belong to the straight line

- Liv1234

Let me guess, it's the first option right?

- Liv1234

Or is it another one?

- Michele_Laino

please wait, first we have to write the equation of the straight line

- Liv1234

Okay, so how would we do that?

- Liv1234

I know that the line is at -6 and 5

- Michele_Laino

your line passes at points (-5,0) and (0,6), so its equation is:
\[\frac{{y - 0}}{{6 - 0}} = \frac{{x + 5}}{{0 + 5}}\]

- Liv1234

It actually passes at -6 and 5 though.

- Michele_Laino

which can be simplified to this form:
\[6x - 5y = - 30\]

- Liv1234

So, what option would I choose?

- Michele_Laino

please wait, I think that your line passes at points (-6,0) and (0,5), so its equation is:
\[5x - 6y = - 30\]

- Liv1234

So, would it be C or D?

- Michele_Laino

so the shaded part of cartesian plane is represented by this inequality:
\[5x - 6y \geqslant - 30\]

- Liv1234

So, D?

- Liv1234

Can you help me with another one please?

- Michele_Laino

more explanation:
the y-coordinate of the points of the shaded region has to be less than the y-coordinate of the point of the straight line, so we can write this:
\[y \leqslant \frac{{5x - 30}}{6}\]

- Liv1234

Can you help me with this question?

##### 1 Attachment

- Liv1234

Please?

- Michele_Laino

we can write this:
\[g\left( x \right) = 5\left( { - 7{x^2}} \right) + 5 = 5f\left( x \right) + 5\]

- Liv1234

I think the answer is D

- Michele_Laino

yes! correct!

- Liv1234

Yay! Can you help me with another one please?

- Michele_Laino

ok!

- Liv1234

##### 1 Attachment

- Liv1234

Okay, that's the question.

- Michele_Laino

the y-coordinates of the new function g(x) have to be six time greater than the respectively y-coordinate of the old function f(x)

- Liv1234

So, would the answer be C?

- Michele_Laino

no, I meant six as a factor

- Liv1234

Oh, so would it be A? Because it has multiplication?

- Michele_Laino

correct!

- Liv1234

Can you help me with another one?

- Michele_Laino

please wait, since another student has requested my help

- Liv1234

It's a quick question though

- Michele_Laino

ok!

- Liv1234

##### 1 Attachment

- Michele_Laino

-7 means shifting down by 7 units
furthermore, as we can see in your first exercise, x+3 means shifting to left

- Liv1234

So, would it be C?

- Liv1234

Oh, wait, B?

- Michele_Laino

no, since the vertical shift is by 7 units down and the horizontal shift is by 3 units left

- Liv1234

So, B?

- Michele_Laino

correct! it is option B

- Liv1234

Do you think you could help me with another?

- Michele_Laino

yes! please wait

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