Asked by Patricia
For each function,
i) determine the corresponding reciprocal
function, y =1/f(x)
ii) state the non-permissible values of
x and the equation(s) of the vertical
asymptote(s) of the reciprocal function
iii) determine the x-intercepts and the
y-intercept of the reciprocal function
iv) sketch the graphs of y = f(x) and
y = 1/f(x) on the same set of axes
a) f(x) = x^2 - 25
b) f(x) = x^2 - 6x + 5
i) determine the corresponding reciprocal
function, y =1/f(x)
ii) state the non-permissible values of
x and the equation(s) of the vertical
asymptote(s) of the reciprocal function
iii) determine the x-intercepts and the
y-intercept of the reciprocal function
iv) sketch the graphs of y = f(x) and
y = 1/f(x) on the same set of axes
a) f(x) = x^2 - 25
b) f(x) = x^2 - 6x + 5
Answers
Answered by
Reiny
I will do the first, you do the second
f(x) = x^2 - 25
let g(x) = 1/(x^2 - 25)
graphs: I assume you know what the parabola looks like, here is the reciprocal
http://www.wolframalpha.com/input/?i=plot++y+%3D+1%2F%28x%5E2+-+25%29
notice that at the x-intercepts of the first one, namely x=±5, the reciprocal has asymptotes,
so vertical asymptotes at x = ± 5
there are no x-intercepts, (since 1/(x^-25) can never be zero.
for the y-intercept , let x = 0
so 1/(0-25) = -.04 (see graph)
f(x) = x^2 - 25
let g(x) = 1/(x^2 - 25)
graphs: I assume you know what the parabola looks like, here is the reciprocal
http://www.wolframalpha.com/input/?i=plot++y+%3D+1%2F%28x%5E2+-+25%29
notice that at the x-intercepts of the first one, namely x=±5, the reciprocal has asymptotes,
so vertical asymptotes at x = ± 5
there are no x-intercepts, (since 1/(x^-25) can never be zero.
for the y-intercept , let x = 0
so 1/(0-25) = -.04 (see graph)
Answered by
Patricia
Thank you so much, the second one is easy now :)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.