Question 1192675
I asume your model function is {{{y =2.8x/(0.4x^2+ 1.3)}}} .
I do not know how the problem would be solved in a nursing class,
so I am going to assume it is for an algebra, pre-calculus, or calculus class.
If you are expected to use a graphing calculator, it should not be difficult.
I am going to assume you are not expected to use a graphing calculator.
 
TO FIND y VALUES:
Plugging x values into that formula, I find
For {{{x=5}}} , {{{y=2.8*5/(0.4*5^2+ 1.3)=14/(0.4*25+13)=14/(10+1.3)=14/11.3=1.238938}}} that I would round to {{{highlight(y=1.24)}}}
For {{{x=10}}} , I find {{{y=0.677966}}} (rounded as {{{highlight(y=0.68)}}} )
For {{{x=20}}} , I find {{{y=0.347179}}} (rounded as {{{highlight(y=0.35)}}} )
For {{{x=30}}} , I find {{{y=0.232494}}} (rounded as {{{highlight(y=0.23)}}} )
For {{{x=60}}} , I find {{{y=0.116561}}} (rounded as {{{highlight(y=0.12)}}} )
 
IMPORTANT NOTE:
That function should be entered into a computer program or a calculator as
y = 2.8x/(0.4x^2+ 1.3)
The parentheses are needed.
If you do not put those parentheses in the formula, it will be calculated is
{{{y}}}{{{"="}}}{{{2.8x/0.4x^2}}}{{{"+"}}}{{{1.3}}} , which does not make sense for concentration in the bloodstream after one dosing.
 
TO FIND x VALUES:
You may be expected to "solve for x" in the formula.
Multiplying times {{{(0.4x^2+ 1.3)}}} both sides of the equal sign in {{{y =2.8x/(0.4x^2+ 1.3)}}} ,
You get {{{(0.4x^2+ 1.3)y=2.8x}}} --> {{{0.4yx^2+1.3y=2.8x}}} --> {{{0.4yx^2-2.8x+1.3y=0}}} .
That is a quadratic equation in {{{x}}}, with coefficients {{{a=0.4y}}} {{{b=-2.8}}} and {{{c=1.3y}}} .
Using the quadratic formula {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
to solve for {{{x}}} ,
we get the formula to find {{{x}}} as a function of {{{y}}} ,
{{{x = (2.8 +- sqrt( 2.8^2-4*(0.4y)(1.3y)))/(2*0.4y) }}} --> {{{highlight(x = (2.8 + sqrt(7.84-2.08y^2))/"0.8 y")}}}
Plugging the given y values into that formula, I get the following rounded values:
For {{{y=0.4}}} , I find {{{highlight(x=17.31)}}} .
For {{{y=0.2}}} , I find {{{highlight(x=34.91)}}}
For {{{y=0.1}}} , I find {{{highlight(x=69.95)}}}
For {{{y=0.01}}} , I find {{{highlight(x=700)}}}