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3 years ago

Graph f(x)=|x+1|+2. Shows vertex and 4 points.

Mathematics

2 answers:

inn [45]3 years ago

8 0

Answer:

$(-1,2)$

Step-by-step explanation:

The vertex of $f(x)=|x|$ is $(0,0)$

The +1 inside the absolute value horizontally shifts the graph 1 unit to the left.

The+2 outside the absolute value vertically shifts the graph 2 units upward.

So, the vertex of the graph of $f(x)=|x+1|+2$ is $(-1,2)$

Wewaii [24]3 years ago

6 0

Vertex: (-1,2)
Here’s the graph

You might be interested in

Use the four-step process to find f'(x) and then find f'(1), f'(2), and f'(3).

Soloha48 [4]

Step 1: evaluate f(x+h) and f(x)

We have

$f(x+h) = -(x+h)^2+6(x+h)-5 = -(x^2+2xh+h^2)+6x+6h-5$

$= -x^2-2xh-h^2+6x+6h-5$

And, of course,

$f(x)=-x^2+6x-5$

Step 2: evaluate f(x+h)-f(x)

$f(x+h)-f(x)=-x^2-2xh-h^2+6x+6h-5-(-x^2+6x-5)=-2xh-h^2+6h$

Step 3: evaluate (f(x+h)-f(x))/h

$\dfrac{f(x+h)-f(x)}{h}=-2x-h+6$

Step 4: evaluate the limit of step 3 as h->0

$f'(x) = \displaystyle \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h}=-2x+6$

So, we have

$f'(1) = -2\cdot 1+6 = 4,\quad f'(2) = -2\cdot 2+6 = 2,\quad f'(3) = -2\cdot 3+6 = 0$

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=log_%7B8%20x%5E%7B2%7D%20-23x%2B15%7D%282x-2%29%20%5Cleq%200" id="TexFormula1" title="log_{8 x

grandymaker [24]

$\log_{8x^2-23x+15} (2x-2) \leq 0$

The domain:
The number of which the logarithm is taken must be greater than 0.
$2x-2 \ \textgreater \ 0 \\
2x\ \textgreater \ 2 \\
x\ \textgreater \ 1 \\ x \in (1, +\infty)$

The base of the logarithm must be greater than 0 and not equal to 1.
* greater than 0:
$8x^2-23x+15\ \textgreater \ 0 \\ 8x^2-8x-15x+15\ \textgreater \ 0 \\ 8x(x-1)-15(x-1)\ \textgreater \ 0 \\ (8x-15)(x-1)\ \textgreater \ 0 \\ \\ \hbox{the zeros:} \\ 8x-15=0 \ \lor \ x-1=0 \\ 8x=15 \ \lor \ x=1 \\ x=\frac{15}{8} \\ x=1 \frac{7}{8} \\ \\
\hbox{the coefficient of } x^2 \hbox{ is greater than 0 so the parabola op} \hbox{ens upwards} \\
\hbox{the values greater than 0 are between } \pm \infty \hbox{ and the zeros} \\ \\
x \in (-\infty, 1) \cup (1 \frac{7}{8}, +\infty)$

*not equal to 1:
$8x^2-23x+15 \not= 1 \\
8x^2-23x+14 \not= 0 \\
8x^2-16x-7x+14 \not= 0 \\
8x(x-2)-7(x-2) \not= 0 \\
(8x-7)(x-2) \not= 0 \\
8x-7 \not=0 \ \land \ x-2 \not= 0 \\
8x \not= 7 \ \land \ x \not= 2 \\
x \not= \frac{7}{8} \\ x \notin \{\frac{7}{8}, 2 \}$

Sum up all the domain restrictions:
$x \in (1, +\infty) \ \land \ x \in (-\infty, 1) \cup (1 \frac{7}{8}, +\infty) \ \land \ x \notin \{ \frac{7}{8}, 2 \} \\ \Downarrow \\
x \in (1 \frac{7}{8}, 2) \cup (2, +\infty)
$

The solution:
$\log_{8x^2-23x+15} (2x-2) \leq 0 \\ \\
\overline{\hbox{convert 0 to the logarithm to base } 8x^2-23x+15} \\
\Downarrow \\
\underline{(8x^2-23x+15)^0=1 \hbox{ so } 0=\log_{8x^2-23x+15} 1 \ \ \ \ \ \ \ }
\\ \\
\log_{8x^2-23x+15} (2x-2) \leq \log_{8x^2-23x+15} 1$

Now if the base of the logarithm is less than 1, then you need to flip the sign when solving the inequality. If it's greater than 1, the sign remains the same.

* if the base is less than 1:
$8x^2-23x+15 \ \textless \ 1 \\
8x^2-23x+14 \ \textless \ 0 \\ \\
\hbox{the zeros have already been calculated: they are } x=\frac{7}{8} \hbox{ and } x=2 \\
\hbox{the coefficient of } x^2 \hbox{ is greater than 0 so the parabola ope} \hbox{ns upwards} \\
\hbox{the values less than 0 are between the zeros} \\ \\
x \in (\frac{7}{8}, 2) \\ \\
\hbox{including the domain:} \\
x \in (\frac{7}{8}, 2) \ \land \ x \in (1 \frac{7}{8}, 2) \cup (2, +\infty) \\ \Downarrow \\ x \in (1 \frac{7}{8} , 2)$

The inequality:
$\log_{8x^2-23x+15} (2x-2) \leq \log_{8x^2-23x+15} 1 \ \ \ \ \ \ \ |\hbox{flip the sign} \\ 2x-2 \geq 1 \\ 2x \geq 3 \\ x \geq \frac{3}{2} \\ x \geq 1 \frac{1}{2} \\ x \in [1 \frac{1}{2}, +\infty) \\ \\ \hbox{including the condition that the base is less than 1:} \\ x \in [1 \frac{1}{2}, +\infty) \ \land \x \in (1 \frac{7}{8} , 2) \\ \Downarrow \\ x \in (1 \frac{7}{8}, 2)$

* if the base is greater than 1:
$8x^2-23x+15 \ \textgreater \ 1 \\ 8x^2-23x+14 \ \textgreater \ 0 \\ \\ \hbox{the zeros have already been calculated: they are } x=\frac{7}{8} \hbox{ and } x=2 \\ \hbox{the coefficient of } x^2 \hbox{ is greater than 0 so the parabola ope} \hbox{ns upwards} \\ \hbox{the values greater than 0 are between } \pm \infty \hbox{ and the zeros}$

$x \in (-\infty, \frac{7}{8}) \cup (2, +\infty) \\ \\ \hbox{including the domain:} \\ x \in (-\infty, \frac{7}{8}) \cup (2, +\infty) \ \land \ x \in (1 \frac{7}{8}, 2) \cup (2, +\infty) \\ \Downarrow \\ x \in (2, \infty)$

The inequality:
$\log_{8x^2-23x+15} (2x-2) \leq \log_{8x^2-23x+15} 1 \ \ \ \ \ \ \ |\hbox{the sign remains the same} \\ 2x-2 \leq 1 \\ 2x \leq 3 \\ x \leq \frac{3}{2} \\ x \leq 1 \frac{1}{2} \\ x \in (-\infty, 1 \frac{1}{2}] \\ \\ \hbox{including the condition that the base is greater than 1:} \\ x \in (-\infty, 1 \frac{1}{2}] \ \land \ x \in (2, \infty) \\ \Downarrow \\ x \in \emptyset$

Sum up both solutions:
$x \in (1 \frac{7}{8}, 2) \ \lor \ x \in \emptyset \\ \Downarrow \\
x \in (1 \frac{7}{8}, 2)$

The final answer is:
$\boxed{x \in (1 \frac{7}{8}, 2)}$

5 0

3 years ago

∠1, ∠2, and ∠3 form a straight line. If m∠3 is six more than twice m∠1 and m∠2 is 27 less than m∠3, find m∠2.

Anarel [89]

Answer:

m<2 = 57

Step-by-step explanation:

Since they are all in a straight line, the sum of the three angles = 180

Let m ∠3 = x

Thus;

2m<1 = x - 6

m<1 = (x-6)/2

m<2 = x-27

Adding all

x + x-27 + (x-6)/2 = 180

Multiply through by 2

2x + 2x -54 + x -6 = 360

5x -60 = 360

5x = 360 + 60

5x = 420

x = 420/5

x = 84

But m<2 = x -27

m<2 = 84 -27 = 57

5 0

4 years ago

What is the quotient of -8a^8b^-2/10a^-4b^-10 in simplified form? Assume a=0 b=0

Alja [10]

Are you sure that both a and b are zero?
I doubt it.

<span>-8a^8b^-2/10a^-4b^-10
should surely be written as a fraction for clarity:

</span><span>-8a^8b^-2
----------------- (at least this is how I interpret your question)
10a^-4b^-10

-8a^8 divided by 10a^(-4) comes out to (-8/10)a^12, and

b^(-2) divided by b^(-10) comes out to b^8

so that your final quotient is (-4/5)(a^12)(b^8).</span>

6 0

3 years ago

Read 2 more answers

Nancy worked thirty-six and three-tenths hours each week How many hours did she work in forty-seven weeks

Annette [7]

Answer:

I believe the answer would be 1,706.1

Step-by-step explanation:

Since the problem is a fraction, it would be easier to covert it to a decimal. This is made simple since the fraction is out of ten. Now, the number should look like this: 36.3

Now let's do long division.

Now we have answer of 1706.1

We have one more step. Since there is one number behind the decimal point in 36.3, we have to put one number behind the decimal point in our answer.

As you can see in the last picture, we have 1,706.1.

I hope you understand after I have explained. If you have any other troubles, tell me.

7 0

3 years ago