For the following function
1 answer:
Answer:
(a) f(4) = 11
(b) f(-1) = 211
(c) f(a) = 5a² -55a +151
(d) f(2/m) = (151m² -110m +20)/m²
(e) x = 5 or x = 6
Step-by-step explanation:
A graphing calculator can help with function evaluation. Sometimes numerical evaluation is easier if the function is written in Horner Form:
f(x) = (5x -55)x +151
(a) f(4) = (5·4 -55)4 +151 = -35·4 +151 = -140 +151 = 11
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(b) f(-1) = (5(-1)-55)(-1) +151 = 60 +151 = 211
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(c) Replace x with a:
f(a) = 5a² -55a +151
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(d) Replace x with 2/m; simplify.
f(2/m) = 5(2/m)² -55(2/m) +151 = 20/m² -110m +151
Factoring out 1/m², we have ...
f(2/m) = (151m² -110m +20)/m²
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(e) Solving for x when f(x) = 1, we have ...
5x² -55x +151 = 1
5x² -55x +150 = 0 . . . . subtract 1
x² -11x +30 = 0 . . . . . . . divide by 5
(x -5)(x -6) = 0 . . . . . . . . factor
Values of x that make the factors (and their product) zero are ...
x = 5, x = 6 . . . . values of x such that f(x) = 1
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Part (a)
The domain is the set of allowed x inputs of a function.
The graph shows that x = 0 is not allowed because of the vertical asymptote located here. It seems like any other x value is fine though.
<h3>Domain: set of all real numbers butTo write this in interval notation, we can say which is the result of poking a hole at 0 on the real number line.
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The range deals with the y values. The graph makes it seem like it stretches on forever in both up and down directions. If this is the case, then the range is the set of all real numbers.
<h3>Range: Set of all real numbers</h3>In interval notation, we would say which is almost identical to the interval notation of the domain, except this time of course we aren't poking at hole at 0.
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Part (b)
<h3>The x intercepts are x = -4 and x = 4</h3>We can compact that to the notation
These are the locations where the blue hyperbolic curve crosses the x axis.
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Part (c)
<h3>Answer: There aren't any horizontal asymptotes in this graph.</h3>Reason: The presence of an oblique asymptote cancels out any potential for a horizontal asymptote.
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Part (d)
The vertical asymptote is located at x = 0, so the equation of the vertical asymptote is naturally x = 0. Every point on the vertical dashed line has an x coordinate of zero. The y coordinate can be anything you want.
<h3>Answer: x = 0 is the vertical asymptote</h3>=======================================================
Part (e)
The oblique or slant asymptote is the diagonal dashed line.
It goes through (0,0) and (2,6)
The equation of the line through those points is y = 3x
If you were to zoom out on the graph (if possible), then you should notice the branches of the hyperbola stretch forever upward but they slowly should approach the "fencing" that is y = 3x. The same goes for the vertical asymptote as well of course.
<h3>Answer: Oblique asymptote is y = 3x</h3>