Finding the set of values of y for which there are no points on a curve

Question

A curve has the equation:

$$y=2x-5+ \frac{18}{x+4}$$

Find the set of values of $y$ for which there are no point on the curve.

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What I did:

Found the maximum and minimum points:

$$\frac{dy}{dx} = 2- \frac{8}{(x+4)^2} = 0 \implies x=-1 , x= -7$$

Found the corresponding $y$ and the nature of these turning points, the results are:

$(-1,-5)$ minimum point

$(-7,-17)$ maximum point

So from my results I concluded:

$-17\leq y \leq -5$

But this is wrong.

Please Help me find my mistake. And mainly

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My question:

Is there a quicker way to do find the set of values? If yes, How?.

Answer 1

Your $y$-coordinate calculations are not correct:

$f(-1)=2(-1)-5+\frac{18}{-1+4}=-1\ne-5$

$f(-7)=2(-7)-5+\frac{18}{-7+4}=-25\ne-17$

Answer 2

for $x=-1$ we get $-2-5+18/3=-7+6=-1$ and for $x=-7$ we obtain $-19-6=-25$

Answer 3

You could use discriminant theory. Consider the equation as a quadratic in x, and find the set of values of y for which there are no real roots

Answer 4

Multiply across and form a quadratic in x. Identify the coefficients which are in terms of y, and set the discriminant to be negative, giving you an inequality to solve for y.