Question

How do you use end behaviour, zeroes, y intercepts to sketch the graph of $ f\left( x \right)=\left( x-4 \right)\left( x-1 \right)\left( x+3 \right) $ ?

Answer 1

Hint: We first find the intercepts of the given function $ f\left( x \right)=\left( x-4 \right)\left( x-1 \right)\left( x+3 \right) $ which also gives us the roots of the function. Then we use the differentiation to find the extremum points of the function and draw the graph.

Complete step-by-step answer:
We need to find the zeros, y intercepts of the curve $ f\left( x \right)=\left( x-4 \right)\left( x-1 \right)\left( x+3 \right) $ .
Here the zeroes mean the x intercepts or the roots of the polynomial.
We put the value of $ f\left( x \right)=0 $ and get $ \left( x-4 \right)\left( x-1 \right)\left( x+3 \right)=0 $ which gives the roots as
 $ x=-3,1,4 $ . The points are $ \left( -3,0 \right),\left( 1,0 \right),\left( 4,0 \right) $ .
To find the y intercepts we put the value of $ x=0 $ and get $ f\left( x \right)=\left( 0-4 \right)\left( 0-1 \right)\left( 0+3 \right)=12 $ which gives the value as the intersecting point of $ \left( 0,12 \right) $ .
Therefore, to find the extremum points we have to find the first and second order derivatives.
Extremum points in a curve have slope value 0.
The slope of the function $ f\left( x \right)=\left( x-4 \right)\left( x-1 \right)\left( x+3 \right) $ can be found from the derivative of the function $ {{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left[ f\left( x \right) \right] $ .
We differentiate both sides of the function $ f\left( x \right)=\left( x-4 \right)\left( x-1 \right)\left( x+3 \right) $ with respect to $ x $ .
 $ \begin{align}
  & f\left( x \right)=\left( x-4 \right)\left( x-1 \right)\left( x+3 \right)={{x}^{3}}-2{{x}^{2}}-11x+12 \\
 & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left[ f\left( x \right) \right]=3{{x}^{2}}-4x-11 \\
\end{align} $ .
To find the $ x $ coordinates of the extremum point we take $ 3{{x}^{2}}-4x-11=0 $ .
In the given equation we have $ 3{{x}^{2}}-4x-11=0 $ . The values of $ a,b,c $ is $ 3,-4,-11 $ respectively.
We put the values and get $ x $ as $ x=\dfrac{4\pm \sqrt{{{4}^{2}}-4\times 3\times \left( -11 \right)}}{2\times 3}=\dfrac{4\pm \sqrt{148}}{6}=\dfrac{2\pm \sqrt{37}}{3} $ .
Therefore, from the value of the $ x $ coordinates of the extremum points, we find their $ y $ coordinates.
Therefore, the extremum points are $ x=\dfrac{2\pm \sqrt{37}}{3} $ .

Note: We need to remember that the curve changes its direction on the extremum points only. Other that that the function $ f\left( x \right)=\left( x-4 \right)\left( x-1 \right)\left( x+3 \right) $ is an increasing function in the range of $ x\in \left[ 4,\infty \right) $ and decreasing function in the range of $ x\in \left( -\infty ,-3 \right] $ .