I will do the first example, by explaining how to graph it. Then you can do the second example.
[tex]f(x) = x^3-10x^2+28x-24[/tex]
We start by determining the the x-intercepts, by letting y = 0:
[tex]x^3-10x^2+28x-24=0,\\\left(x-2\right)^2\left(x-6\right)=0,\\x=2,\:x=6[/tex]
The x-intercepts of the graph are (2, 0) and (6, 0) respectively. Now we determine the y-intercepts by letting x = 0:
[tex]f\left(x\right)=\left(0\right)^3-10\left(0\right)^2+28\left(0\right)-24 = - 24[/tex]
So the y-intercept is (0, -24). But remember we need one more point, as there has to be a transition between the x-intercepts. This is the minimum of the graph. To determine this, we take the derivative of the function, and then equate it to 0 to find the "critical points." :
[tex]\frac{d}{dx}\left(x^3-10x^2+28x-24\right)\\\\ =>\frac{d}{dx}\left(x^3\right)-\frac{d}{dx}\left(10x^2\right)+\frac{d}{dx}\left(28x\right)-\frac{d}{dx}\left(24\right)\\\\=> 3x^2-20x+28-0\\=> 3x^2-20x+28[/tex]
Now we equate this to 0, and apply the quadratic equation. In this case we can actually factor it :
[tex]3x^2-20x+28=0,\\\left(x-2\right)\left(3x-14\right)=0,\\\\x=2,\:x=\frac{14}{3}[/tex]
x = 2 is our maximum, or our x-intercept (2, 0). Therefore we have to take 14/3 as our minimum, and determine the y-value:
[tex]f\left(x\right)=\left(\frac{14}{3}\right)^3-10\left(\frac{14}{3}\right)^2+28\left(\frac{14}{3}\right)-24\\\\= -\frac{256}{27}[/tex]
Minimum: (14/3, -256/27)
Graph is in the attachment below -
