Answer
(a) $-\frac{1}{2}a^{-3/2}$
(b) For point (1,1)
$y-1=-\frac{1}{2}(x-1)$
For point (4,$\frac{1}{2}$)
$y-\frac{1}{2}=-\frac{1}{16}(x-4)$
(c) red: $y=\frac{1}{\sqrt x}$
purple: $y=-\frac{1}{2}x+\frac{3}{2}$
green: $y=-\frac{1}{16}x+\frac{3}{4}$
Work Step by Step
(a) $y=\frac{1}{\sqrt x}$ at the point where $x=a$
Step 1: The question is asking for the slope of the tangent line so we have to find the first derivative. We can find the first derivative by using the power rule.
$y=\frac{1}{\sqrt x}$
$y=x^{-1/2}$
$y'=-\frac{1}{2}x^{-3/2}$
Step 2: Plug in a wherever you see an x in the equation.
$-\frac{1}{2}a^{-3/2}$
(b) points (1,1) and (4,$\frac{1}{2}$)
Step 1: Plug in the given x values into $y'=-\frac{1}{2}x^{-3/2}$
$y'(1)=-\frac{1}{2}(1)^{-3/2}$
$y'(1)=-\frac{1}{2}$
$y'(4)=-\frac{1}{2}(4)^{-3/2}$
$y'(4)=-\frac{1}{16}$
Step 2: Now that we have calculated the slope of the tangent lines, we can write the equation of the tangent line in point slope form.
For point (1,1)
$y-1=-\frac{1}{2}(x-1)$
For point (4,$\frac{1}{2}$)
$y-\frac{1}{2}=-\frac{1}{16}(x-4)$
Step 3: Rewrite the equations into standard form.
For point (1,1)
$y=-\frac{1}{2}x+\frac{3}{2}$
For point (4,$\frac{1}{2}$)
$y=-\frac{1}{16}x+\frac{3}{4}$
(c) Plug the original function and equations of the tangent lines into a graphing calculator or graph by hand.
red: $y=\frac{1}{\sqrt x}$
purple: $y=-\frac{1}{2}x+\frac{3}{2}$
green: $y=-\frac{1}{16}x+\frac{3}{4}$
