Quick little explanation of easy equation of tangent like

g(x)  =  \(5-\sqrt{4-x}\)

When  x = 3 , the slope of the curve  \(=\,\lim\limits_{h\to0}\frac{g(3+h)-g(3)}{h} \\~\\ =\,\lim\limits_{h\to0}\frac{(5-\sqrt{4-(3+h)})-(5-\sqrt{4-3})}{h} \\~\\ =\,\lim\limits_{h\to0}\frac{5-\sqrt{4-3-h}-4}{h} \\~\\ =\,\lim\limits_{h\to0}\frac{1-\sqrt{1-h}}{h} \\~\\ =\,\lim\limits_{h\to0}(\,\frac{1-\sqrt{1-h}}{h}\,)\,(\,\frac{1+\sqrt{1-h}}{1+\sqrt{1-h}}\,) \\~\\ =\,\lim\limits_{h\to0}\frac{1-(1-h)}{h+h\sqrt{1-h}} \\~\\ =\,\lim\limits_{h\to0}\frac{h}{h+h\sqrt{1-h}} \\~\\ =\,\lim\limits_{h\to0}\frac{1}{1+\sqrt{1-h}} \\~\\ =\,\frac{1}{1+\sqrt{1-0}} \\~\\ =\,\frac12\)

Look at the graph and see that it does look like that line has a slope of \(\frac12\) .

We want an equation of a line that has a slope of  \(\frac12\)  and passes through the point  ( 3, g(3) ) .

g(3)  =  \(5-\sqrt{4-3}\)  =  5 - 1  =  4

So the equation of our line in point - slope form is

y - 4  =  \(\frac12\)(x - 3)

And in slope - intercept form, this is...     y  =  \(\frac12\)x + \(\frac52\)

Very nice, hectictar........!!!!