The function f(x) is defined as,
ormula1" title="f(x) = 3 + \frac{1}{2x-5}" alt="f(x) = 3 + \frac{1}{2x-5}" align="absmiddle" class="latex-formula">, x ≠ 
(a) Sketch the curve of f for -5 ≤ x ≤ 5, showing the asymptotes.
(b) Using your sketch, write down
(i) the equation of each asymptote;
(ii) the value of the x-intercept;
(iii) the value of the y-intercept.
Image attached.
(a) Sketch the curve of f for -5 ≤ x ≤ 5, showing the asymptotes.
(b) Using your sketch, write down
(i) the equation of each asymptote;
(ii) the value of the x-intercept;
(iii) the value of the y-intercept.
Image attached.
1 answer:
Answer:
a) look at the figure
b) i) x = 2.5 , y = 3
ii) x-intercept is 2.333
iii) y-intercept is 2.8
Step-by-step explanation:
a) The points of the graph f(x) = 3+1/2x - 5 are:
f(-5) = 2.9333
f(-4) = 2.923
f(-3) = 2.909
f(-2) = 2.888
f(-1) = 2.857
f(0) = 2.8
f(1) = 2.666
f(2) = 2
f(3) = 4
f(4) = 3.333
f(5) = 3.2
b)
i) to find the vertical asymptotic put the denominator = 0
2x - 5 = 0 ⇒ 2x = 5 ⇒ x = 5 ÷ 2 = 2.5
∴ The equation of the vertical asymptotic is x = 2.5
To find the horizontal asymptotic look at the degree of the numerator and denominator
∵ they are equal f(x) = (6x -14)/(2x - 5) ⇒ 6x ÷ 2x = 3
∴ The equation of the horizontal asymptotic is y = 3
ii) the value of x-intercept means put f(x) = 0
∴3 + 1/2x - 5 = 0 ⇒ 1/2x - 5 = -3 ⇒ -6x + 15 = 1 ⇒ 6x = 14
x = 14/6 = 2.333
iii) The value of y-intercept means x = 0
∴ f(x) = 3 + 1/0 - 5 = 3 + (-0.2) = 2.8
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<h2>Hello!</h2>
The answer is:
A multiple of a number is the repeated sum of itself, for example:
Multiple of the number 1 are: 1, 2, 3, 4, 5 and so...
Multiple of the number 3 are: 3, 6, 9, 12, 15 and so...
From the example we have:
Multiple of the number 3 are: Itself, (3+3), (3+3+3), (3+3+3+3), (3+3+3+3+3) and so...
So,
Multiple of are:
Have a nice day!
Changing the equation into slope form: 
, where 
is the slope [gradient] and 
is the y-intercept.
The gradient is
and y-intercept is at 
Graphing using slope-intercept method:
a) The slope is a negative slope. The line will go 'down hill'
b) The line must pass the point (0, 490)
c) The line will intercept the x-axis at y = 0
So, x-intercept is at (735, 0)
The graph of this function is shown below. The intercepts are labelled at:
y = 490
x = 735
-----------------------------------------------------------------------------------------------------------
Next month's profit equation
Rewriting this into slope-equation form
The gradient,, equals to
The y-intercept,, equals to 531
The equation still has the same gradient with last month's profit equation but different y-intercept.
-------------------------------------------------------------------------------------------------------------
A linear graph show points of (0, 300) and (450, 0)
We work out the slope:
Y-intercept at x = 0, so it's at y = 300
Equation
The gradient is
Graphing
a) The slope is a negative slope. The line will go 'down hill'
b) The line must pass the point (0, 490)
c) The line will intercept the x-axis at y = 0
So, x-intercept is at (735, 0)
The graph of this function is shown below. The intercepts are labelled at:
y = 490
x = 735
-----------------------------------------------------------------------------------------------------------
Next month's profit equation
Rewriting this into slope-equation form
The gradient,
The y-intercept,
The equation still has the same gradient with last month's profit equation but different y-intercept.
-------------------------------------------------------------------------------------------------------------
A linear graph show points of (0, 300) and (450, 0)
We work out the slope:
Y-intercept at x = 0, so it's at y = 300
Equation
The two dot plots are missing, so i have attached it.
Answer:
The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles
Step-by-step explanation:
From the attached image, we are told to compare the means for each plot to the nearest tenth.
Mean = Σx/n
Now, from the image, total number of miles run by the 14 students at the beginning of the school year is;
(1 × 7) + (2 × 8) + (4 × 9) + (4 × 10) + (2 × 11) + (1 × 12) = 133
Mean of miles run at the beginning of the school year = 133/14 = 9.5 miles
Again, from the table, total miles run at the end of the school year = (2 × 8) + (2 × 9) + (4 × 10) + (3 × 11) + (3 × 12) = 143
Mean of miles run at the end of the school year = 143/14 = 10.2 miles
Thus;
The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles
