#### Answer

(a) $8a-6a^2$
(b)For point (1,5)
$y=2x+3$
For point (2,3)
$y=-8x+19$
(c) red = $y=3+4x^2-2x^3$
blue = $y=2x+3$
green = $y=-8x+19$

#### Work Step by Step

(a) $y=3+4x^2-2x^3$ 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.
$y'=8x-6x^2$
Step 2: Plug in $a$ wherever you see an $x$ in the equation.
$8a-6a^2$
(b) points (1,5) and (2.3)
Step 1: Plug in the given x values into $y'=8x-6x^2$.
$y'(1)=8(1)-6(1)^2$
$y'(1)=2$
$y'(2)=8(2)-6(2)^2$
$y'(2)=-8$
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 (1,5)
$y-5=2(x-1)$
For point (2,3)
$y-3=-8(x-2)$
Step 3: Rewrite the equations into standard form.
For point (1,5)
$y=2x+3$
For point (2,3)
$y=-8x+19$
(c) Plug the original function and equations of the tangent lines into a graphing calculator or graph by hand.
red = $y=3+4x^2-2x^3$
blue = $y=2x+3$
green = $y=-8x+19$