Show that y is always positive for all real values of x hence sketch the curve y = f(x) showing clearly the asymptotes, intercepts, with coordinate axes and the behaviour of the curve as it approaches its asymptote and the behaviour of the curve as it approaches itsSketch the curve y = f(x), showing clearly theshow that isasymptotes, intercepts with the coordinate axes3.13) Given that y 2-x² - 2x+4always positive for all real values of x.asymptotes, intercepts with the coordinate avenasymptote(s).

Answered: 3 show that is Given that y = 2- x²…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.5: Rational Functions

Problem 54E

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Show that y is always positive for all real values of x hence sketch the curve y = f(x) showing clearly the asymptotes, intercepts, with coordinate axes and the behaviour of the curve as it approaches its asymptote

and the behaviour of the curve as it approaches its
Sketch the curve y = f(x), showing clearly the
show that is
asymptotes, intercepts with the coordinate axes
3.
13) Given that y 2-
x² - 2x+4
always positive for all real values of x.
asymptotes, intercepts with the coordinate aven
asymptote(s).

Expert Solution

Step 1

The set of values of the dependent variable for which a function is defined.

The range of the given function is given as:

The function range is the combined domain of the inverse function. first find the inverse function of the given function:

$\begin{array}{l} Inverse of 2 - \frac{3}{x^{2} - 2 x + 4} \\ A function g is the inverse of function f if for y = f (x), x = g (y) \\ y = 2 - \frac{3}{x^{2} - 2 x + 4} \\ Replace x with y : \\ x = 2 - \frac{3}{y^{2} - 2 y + 4} \{Solve for y\} \\ x - 2 = - \frac{3}{y^{2} - 2 y + 4} \\ x (y^{2} - 2 y + 4) - 2 (y^{2} - 2 y + 4) = - 3 \\ y = \frac{- 4 + 2 x + \sqrt{- 12 x^{2} + 36 x - 24}}{2 (x - 2)} \\ and y = \frac{- 4 + 2 x - \sqrt{- 12 x^{2} + 36 x - 24}}{2 (x - 2)} \end{array}$

Step 2

Now, the domain of the inverse function is the range of the given function:

$\begin{array}{l} Domain y = \frac{- 4 + 2 x + \sqrt{- 12 x^{2} + 36 x - 24}}{2 (x - 2)} \\ The domain of a function is the set of input or argument values for which the fucntion is real and defined. \\ The non negative values for radicals 1 \leq x \leq 2 \\ But the function is undefined at x = 2 \\ Combine real regions and undefined points for final function domain \\ 1 \leq x < 2 \\ and similarily domain of y = \frac{- 4 + 2 x - \sqrt{- 12 x^{2} + 36 x - 24}}{2 (x - 2)} is 1 \leq x < 2 \\ Thus, Range of 2 - \frac{3}{x^{2} - 2 x + 4} : [\begin{matrix} Solution : & 1 \leq f (x) < 2 \\ Interval Notation : & [1, 2) \end{matrix}] \end{array}$

Step 3

Asymptotes and intercepts:

The x-intercept is the point at which the graph crosses the x-axis or the y-coordinate is zero.

The y-intercept is the point at the x-coordinate is zero.

$\begin{array}{l} x - intercept of y = 2 - \frac{3}{x^{2} - 2 x + 4} : N o n e \\ Put y = 0 into the equation and solve the equation: \\ 0 = 2 - \frac{3}{x^{2} - 2 x + 4} \\ \Rightarrow - 2 (x^{2} - 2 x + 4) = - 3 \\ x = 1 - i \sqrt{\frac{3}{2}}, x = 1 + i \sqrt{\frac{3}{2}} \{No real x, No x - intercept\} \\ y - intercept of 2 - \frac{3}{x^{2} - 2 x + 4} : (0, \frac{5}{4}) \\ When x = 0 \\ y = 2 - \frac{3}{{(0)}^{2} - 2 (0) + 4} = 2 - \frac{3}{4} = \frac{5}{4} \end{array}$

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