Shayan A.

asked • 04/10/17

Find the set of values of k for which f(x)= 3x^2 -5x -k is greater than 1 for all real x

This is a year 11 methods question about applying function notation.
Could someones please explain the question and then give me a working out for it?

Prentice D.

Ok so this question is asking for what values for k such that the graph never goes below 1. Since this is a hyperbolic function of the form f(x)=ax2 +bx+c (happy face graph due to positive a value), the function will have a lowest point when the derivative of this function is equal to zero. So,
f(x) = 3x2 -5x -k
f'(x) = 6x -5   (notice that the derivative does not depend on k, so the lowest point of the graph will have the same x value no matter the k value).
now sub in f'(x) = 0
0 = 6x-5
Now from the original function sub in f(x) = 1, and x=5/6
1 = 3*(5/6)^2 -5(5/6) -k
k=-3.08 (approximately)
Now the k value will only lower or increase the height of the graph, so increasing the value for k value will put the lowest part on the graph above 1, while decreasing 1 will make the lowest part on the graph less than 1.
So therefore the value for k must be k>-3.08, so that the graph is greater than 1 for all real x.


Shayan A.

Thank you for your response,
It makes sense using the derivative (I had some personal practice on it before) but we havent officially been taught yet. So I was wondering if there is another way of solving this question.


1 Expert Answer


Roman C. answered • 04/10/17

Masters of Education Graduate with Mathematics Expertise

Shayan A.

Thank you for your response,
I was just wondering where does the y=-(discriminant)/4a comes from?
Whats the reason that it works?


Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.


Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.