If the line has to pass through $(0,b)$, $(3,2)$ and $(a,0)$ it has to satisfy the following equation:
$$
y-0=\dfrac{2-0}{3-a}(x-a)
$$
Observe that when $x=0$, $y=-\dfrac{2}{3-a}a$, that is $b=-\dfrac{2}{3-a}a$
Now if both $a,b$ are non-negative, then certainly the line will have negative slope (why?). Now the area $A(a,b)$ of the triangle formed by the $x$-axis, the $y$-axis and the line, is given by $A(a,b)=\dfrac{1}{2}ab$. By replacing the value of $b$ in this equation we obtain a formula for the area of the triangle as a function on $a$, $A(a)=-\dfrac{1}{2}a\dfrac{2}{3-a}a=\dfrac{a^2}{a-3}$. Now if you want to know for what value of $a$ this function has a maximum or a minimum, differential calculus is of great help.
$$
A'(a)=\dfrac{2a(a-3)-a^2}{(a-3)^2}=-\dfrac{a^2-6a}{(a-3)^2}
$$
To find the critical points of this function we have to find the zeroes of $A'(a)$, that is, we need to find the solutions of the following equation:
$$
-\dfrac{a^2-6a}{(a-3)^2}=0
$$
In this case the zeroes will come from the zeroes of the numerator that is $a^2-6a=0$, but those are $a=0,\quad a=6$, We have to reject $a=0$ since it does not produces a triangle. The answer is $a=6$. So the value of the area will be $A(6)=\dfrac{1}{2}\dfrac{6^2}{(6-3)^2}=2$. In order to find out whether this is a maximum or a minimum you could use the second derivative criteria:
$$
A''(a)=-\dfrac{4(a-3)^3(a^2-6a)-(a-3)^4(2a-6)}{(a-3)^4}=-\dfrac{2(a-3)^3(2a^2-12a-(a-3)(a-3)}{(a-3)^4}=-\dfrac{2a^2-12a-a^2+6a-9}{(a-3)^3}=-\dfrac{a^2-6a-9}{(a-3)^3}
$$
So $A''(6)=-\dfrac{6^2-6(6)-9}{(6-3)^3}=\dfrac{9}{27}=\dfrac{1}{3}>0$, the area is a minimum.
Hint: Let the slope of the line be $m$, then the point-slope equation for a line gives
$$y - 4 = m(x - 3)$$
Now find the $x$-intercept (plug in $y = 0$ and solve for $x$, your answer will be a formula involving $m$), find the $y$-intercept (plug in $x = 0$), and use these two points to calculate the area of the triangle (area $= \frac{1}{2}$base$\cdot$height). This is again a formula involving $m$. You should now be able to find what value of $m$ minimizes this formula.
Best Answer
Please note I am using notation $m$ for the slope instead of $a$.
If you have two points $A$ and $B$ with $x-$coordinates $x = a$ and $x =b$ ($b \gt a$) on the parabola $y=x^2$, area of parabola under the straight line between these two points can be found as below.
For points on the line segment $AB$, $y -a^2 = \frac{b^2-a^2}{b-a} (x-a) \implies y = (a+b)x - ab$
So the area is $A = \int_a^b ((a+b)x-ab - x^2) \ dx = \frac{1}{6}(b-a)^3$ ...(i)
Now for a line with slope $m$ through point $(-1, 2)$, the equation of the line will be
$y-2 = m(x+1)$ and $x-$coordinates of its intersection with parabola $y=x^2 \ $ is given by,
$x^2 - mx - (m+2) = 0 \implies x = \frac{m \pm \sqrt{m^2+4m+8}}{2}$
Plugging in $(i)$, $A = \frac{1}{6}(m^2+4m+8)^{3/2}$
So if we minimize $(m^2+4m+8)$, we minimize the area. We can rewrite $m^2+4m+8$ as $(m+2)^2+4$ which shows that the minimum area occurs at $m = -2$.