Answer
a. $8a - 6a^{2}$
b. Tangent lines: $y = 2x + 3$ and $y = -8x + 19$
c. Graph
Work Step by Step
$y = 3 + 4x^{2} - 2x^{3}$
$\lim\limits_{h \to 0} \frac{3+4(a+h)^{2}-2(a+h)^{3} - (3 + 4a^{2} - 2a^{3}}{h}$
Expand:
$\lim\limits_{h \to 0} \frac{3+4a^{2} + 8ah + 4h^{2} -2a^{3} - 6a^{2}h -6ah^{2} - 2h^{3} - 3 -4a^{2} + 2a^{3}}{h}$
Simplify:
$\lim\limits_{h \to 0} \frac{8ah + 4h^{2} - 6a^{2}h -6ah^{2} - 2h^{3}}{h}$
Divide by $h$.
$\lim\limits_{h \to 0} \frac{h(8a + 4h - 6a^{2} -6ah - 2h^{2})}{h}$
Cancel $h$.
$\lim\limits_{h \to 0} 8a + 4h - 6a^{2} - 6ah - 2h^{2}$
Replace $h$ to $0$
$\lim\limits_{h \to 0} 8a + 4(0) - 6a^{2} - 6a(0) - 2(0)^{2} = 8a - 6a^{2}$
b. Tangent line
Slope from part a: $8a - 6a^{2}$
The problem gives you the intercepts of $(1,5)$ and $(2,3)$ so we use the value of $x$ to find the tangent lines.
With $(1,5)$:
$m = 8a - 6a^{2}$
$m = 8(1) - 6(1)^{2}$
$m = 8 -6 = 2$
Now replace it in the formula:
$y - y_{1} = m(x-x_{1})$
$y = 2(x-1) + 5$
$y = 2x + 3$
With $(2,3)$:
$m = 8a - 6a^{2}$
$m = 8(2) - 6(2)^{2}$
$m = 16 - 24 = -8$
Now replace it in the formula:
$y - y_{1} = m(x-x_{1})$
$y = -8(x-2) + 3$
$y = -8x + 19$
