Curve Sketching
This lesson is meant to summarize all the major concepts covered from this unit. The original function and its respective derivatives are used to find different pieces of information about the function. We can combine all of this information to help draw an accurate sketch of the original function.
From \(f(x)\):
- \(x\) and \(y\) intercepts
- Domain and Vertical Asymptote
- Horizontal and Oblique Asymptotes
From \(f'(x)\):
- Critical Points
- Max/Min Values
- Intervals of Increase/Decrease
From \(f''(x)\):
- Critical Point Info
- Interval Info
- Inflection Points
- Intervals of Concavity
In order to identify the \(x\)-intercept, we need to set the function (or numerator if it's rational) equal to \(0\). We can then solve for \(x\). If we can't cleanly factor the function, then we can use either the Quadratic Equation or use Synthetic Division to find the roots.
In order to identify the \(y\)-intercept, we can substitute \(x = 0\) and solve for \(y\).
Asymptotes only exist for Rational Functions.
In order to determine the Vertical Asymptote, we can set the denominator equal to \(0\). We can then solve for \(x\). The \(x\)-intercepts cut the \(x\)-axis based on the exponent (ie cut, bounce, squiggle).
Horizontal Asymptotes only exist if the degree of the numerator and denominator are the exact same. If so, then find the ratio of their leading coefficients.
Oblique Asymptotes only exist if the numerator is exactly 1 degree higher than the numerator. If so, then use long division to find its equation.
At the end, find all y-values of all the Special Points outlined above in order to plot them accurately.
In order to determine the Critical Points, we can find the first derivative of the original function, \(f'(x)\). We can then factor this expression and set it equal to 0 to find its roots. If we can't factor this function cleanly, then we can use either the Quadratic Equation or Synthetic Division, similar to finding the \(x\)-intercepts.
In order to identify the Intervals of Change, we can use the First Derivatives Test. From this, we can determine which intervals the original function is increasing and which its decreasing. If the function is increasing (+) then decreasing (-), this indicates a Local Maximum. If the function increases (+) then decreases (-), then that represents a Local Minimum.
In order to identify the Concavity of a function, we can find the Second Derivative of the original function, \(f''(x)\). We can then factor this expression and set it equal to 0 to find its roots. If we can't factor this function cleanly, then we can use either the Quadratic Equation or Synthetic Division, similar to finding the \(x\)-intercepts or Critical Points.
In order to determine the actual Point(s) of Inflection, we can conduct the Second Derivatives Test. From this, we can determine the respective concavities of each interval. If the concavity changes from positive (+) to negative (-) in between intervals, this represents a Point of Inflection.
Enter in a coefficient for quadratic monomial or click on the New Question button for a random value. Then, use the First Principles formula to solve for the derivative.
First, we can factor the function in order to analyze it more easily:
\(p(x) = \cfrac{4-x}{(x+1)^2}\)
Next, we can determine the function's intercepts and asymptotes. We'll start with the \(y\)-intercept:
\(\text{y-int}: y = \cfrac{4-0}{(0+1)^2}\)
\(\text{y-int}: y = \cfrac{4}{1} = 4\)
Then, we can determine the \(x\)-intercept:
\(\text{x-int}: 0 = 4 - x\)
\(\text{x-int}: x = 4\)
After, we can determine the Vertical Asymptote:
\(\text{VA}: x+1\)
\(\text{VA}: x = -1\)
Finally, we can determine the Horizontal Asymptote:
\(\text{HA}: \text{N} < \text{D}\)
\(\text{HA}: 0\)
Therefore, we can determine that the \(y\)-intercept is \(4\) and the \(x\)-intercept is \(4\). We can also determine that the Vertical Asymptote is located at \(x = -1\) and the Horizontal Asymptote is located at \(y=0\).
Next, we can determine the Critical Point(s) by differentiating the function using Quotient Rule:
\(p'(x) = \cfrac{\textcolor{blue}{g(x)}\textcolor{red}{f'(x)} - \textcolor{red}{f(x)}\textcolor{blue}{g'(x)}}{[\textcolor{blue}{g(x)}]^2}\)
\(p'(x) = \cfrac{(\textcolor{blue}{(x+1)^2})(\textcolor{red}{-1}) - (\textcolor{red}{4-x})(\textcolor{blue}{2x + 2})}{[\textcolor{blue}{(x+1)^2}]^2}\)
\(p'(x) = \cfrac{(-1)(x+1)^2 - 2(x + 1)(4-x)}{[(x+1)^2]^2}\)
Then, we can simplify by taking a common factor out of the numerator and cancelling it out of the denominator:
\(p'(x) = \cfrac{(-1)(x+1)^2 - 2(x + 1)(4-x)}{[(x+1)^2]^2}\)
\(p'(x) = \cfrac{\cancel{(x+1)}[(-1)(x+1) - 2(4-x)}{[(x+1)^2]^\cancel{2}}\)
\(p'(x) = \cfrac{-x-1 -8 + 2x}{(x+1)^3}\)
Finally, we can simplify further by collecting like terms:
\(y' = \cfrac{-x-1-8+2x}{(x+1)^3}\)
\(y' = \cfrac{x-9}{(x+1)^3}\)
After, we can determine the critical point(s) by setting the numerator and denominator to \(0\) and solving for \(x\).
We can determine the critical point for the numerator as such:
\(0 = x-9\)
\(x_1 = 9\)
We can determine the critical point for the denominator as such:
\(0 = x+1\)
\(x_2 = -1\)
Subsquently, we can use the First Derivatives Test to determine the Interval(s) of Change:
| Interval | Test \(x\)-value | \(p'(x)\) | Conclusion |
|---|---|---|---|
| \((-\infty,-1)\) | \(-2\) | \(p'(-2)= 11\) | \(p\) is Increasing |
| \((-1, 9)\) | \(4\) | \(p'(4)= -1/25\) | \(p\) is Decreasing |
| \((9, \infty)\) | \(10\) | \(p'(10)= 1/1331\) | \(p\) is Increasing |
From this table, we are able to determine that this function has a minimum point at \(x=9\).
In order to identify the Point(s) of Inflection, we can find the second derivative using Quotient Rule:
\(p''(x) = \cfrac{\textcolor{blue}{g(x)}\textcolor{red}{f'(x)} - \textcolor{red}{f(x)}\textcolor{blue}{g'(x)}}{[\textcolor{blue}{g(x)}]^2}\)
\(p''(x) = \cfrac{\textcolor{blue}{(x+1)^3}(\textcolor{red}{1}) - (\textcolor{red}{x-9})\cdot \textcolor{blue}{3(x+1)^2(1)}}{[\textcolor{blue}{(x+1)^3}]^2}\)
Next, we can simplify this expression by taking a common factor out of the numerator and cancelling out the denominator:
\(p''(x) = \cfrac{(x+1)^3(1) - 3(x-9)(x+1)^2(1)}{(x+1)^6}\)
\(p''(x) = \cfrac{\cancel{(x+1)^2}[(1)(x+1) - 3(1)(x-9)]}{(x+1)^{\cancel{6}4}}\)
\(p''(x) = \cfrac{x+1 - 3x+27}{(x+1)^4}\)
Then, we can simplify further by collecting like terms:
\(p''(x) = \cfrac{x+1-3x+27}{(x+1)^4}\)
\(p''(x) = \cfrac{-2x+28}{(x+1)^4}\)
After, we can determine the potential Point(s) of Inflection by setting the numerator and denominator to \(0\) and solving for \(x\).
We can solve for \(x\) for the numerator as such:
\(0=-2x+28\)
\(2x=28\)
\(x_1=14\)
We can solve for \(x\) for the denominator as such:
\(0=x+1\)
\(x_2=-1\)
Therefore, we can determine that our potential Points of Inflection are \(x=14\) and \(x=-1\).
Finally, we can use the Second Derivatives Test to determine the actual Point of Inflection:
| Interval | Test \(x\)-value | \(p''(x)\) | Conclusion |
|---|---|---|---|
| \((-\infty,-1)\) | \(-2\) | \(p''(-2)= 32\) | \(p\) is Concave Up |
| \((-1, 14)\) | \(7\) | \(p''(7)= 14/4096\) | \(p\) is Concave Up |
| \((14, \infty)\) | \(15\) | \(p''(15)= -2/65536\) | \(p\) is Concave Down |
We can determine that the Point of Inflection lies at \(x = 14\). This is because the sign changes from (+) to (-) before and after this point respectively. In addition, \(x = -1\) doesn't count as it represents a VA.
Given all the information we have gathered, we can now sketch our graph:
First, we can determine the \(y\)-intercept:
\(\text{y-int}: y = (0)^4 -5(0)^3 + (0)^2 + 21(0) - 18\)
\(\text{y-int}: y = -18\)
In order to determine the \(x\)-intercepts, we can use Synthetic Division:
\(\begin{array}{c|rrr}&1&-5&1&21&-18\\1&&1&-4&-3&18\\\hline\\&1&-4&-3&18&0\\\end{array}\)
Therefore, we can factor the equation as such:
\(y = (x-1)(x^3 -4x^2 -3x + 18)\)
We can factor the equation further by using Synthetic Division again:
Therefore, we can factor the equation as such:
We can factor the equation even further:
\(y = (x-1)(x-3)(x^2 -3x + 2x - 6)\)
\(y = (x-1)(x-3)(x(x+2)-3(x+2))\)
\(y = (x-1)(x-3)(x+2)(x-3)\)
\(y = (x-1)(x+2)(x-3)^2\)
Therefore, we can determine that the \(y\)-intercept is \(-18\) and the \(x\)-intercepts are \(1\), \(-2\), and \(3\).
Next, we can determine the Critical Points by finding the first derivative:
\(y' = 4(x^3) 3(-5x^2) + 2(x) + 1(21) - 0(18)\)
\(y' = 4x^3 -15x^2 + 2x + 21\)
In order to determine the Critical Points, we can use Synthetic Division:
\(\begin{array}{c|rrr}&4&-15&2&-21\\-1&&-4&19&21\\\hline\\&4&-19&21&0\\\end{array}\)
Therefore, we can factor the equation as such:
\(y' = (x+1)(4x^2 -19x + 21)\)
We can factor this equation even further to find the additional Critical Points:
\(y' = (x+1)(4x^2 -12x -7x + 21)\)
\(y' = (x+1)[4x(x -3)-7(x -3)]\)
\(y' = (x+1)(x -3)(4x-7)\)
After, we can find the critical points by setting the equation to \(0\) and solving for \(x\).
We can solve for \(x\) for the first factor:
\(0 = x+1\)
\(x_1 = -1\)
We can solve for \(x\) for the second factor:
\(0 = x -3\)
\(x_2 = 3\)
We can solve for \(x\) for the third factor:
\(0 = 4x-7\)
\(4x = 7\)
\(x_3 = \cfrac{7}{4}\)
Therefore, we can determine the Critical Points are \(x = -1, 3, 1.75\).
We can now use the First Derivatives Test in order to identify the Intervals of Change:
| Interval | Test \(x\)-value | \(y'(x)\) | Conclusion |
|---|---|---|---|
| \((-\infty,-1)\) | \(-2\) | \(y'(-2)= -75\) | \(y\) is Decreasing |
| \((-1, 1.75)\) | \(0\) | \(y'(0)= 21\) | \(y\) is Increasing |
| \((1.75, 3)\) | \(2\) | \(y'(2)= -3\) | \(y\) is Decreasing |
| \((3, \infty)\) | \(4\) | \(y'(4)= 45\) | \(y\) is Increasing |
From this table, we are able to determine that this equation has min points at \(x = -1\) and \(x = 3\) and a max point at \(x = 1.75\).
In order to find the Points of Inflection, we can determine the Second Derivative:
\(y' = 4x^3 -15x^2 + 2x + 21\)
\(y'' = 3(4x^2) + 2(-15x) + 2\)
\(y'' = 12x^2 -30x + 2\)
Since we can't factor this equation cleanly, we can use the Quadratic Equation in order to find its factors:
\(x = \cfrac{-\textcolor{green}{b} \pm \sqrt{\textcolor{green}{b}^2-4\textcolor{red}{a}\textcolor{blue}{c}}}{2\textcolor{red}{a}}\)
\(x = \cfrac{-(\textcolor{green}{-30}) \pm \sqrt{(\textcolor{green}{-30})^2-4(\textcolor{red}{12})(\textcolor{blue}{2})}}{2(\textcolor{red}{12})}\)
\(x = \cfrac{30 \pm \sqrt{900-96}}{24}\)
\(x = \cfrac{30 \pm 2\sqrt{201}}{24}\)
Using \(\pm\), we can determine the 2 roots either by adding the 2 values of subtracting the 2 values:
We can determine the first root by adding the values in the numerator:
\(x_1 = \cfrac{30 + 2\sqrt{201}}{24}\)
\(x_1 = 2.43\)
We can determine the second root by subtracting the values in the numerator:
\(x_2 = \cfrac{30 - 2\sqrt{201}}{24}\)
\(x_2 = 0.07\)
Therefore, we can determine that the potential Points of Inflection are at \(x = 0.66, 2.43\).
We can rewrite the Second Derivative equation as such:
Finally, we can use the Second Derivatives Test to determine the actual Point of Inflection:
| Interval | Test \(x\)-value | \(y''(x)\) | Conclusion |
|---|---|---|---|
| \((-\infty,0.07)\) | \(0\) | \(y''(0)= 2\) | \(y\) is Concave Up |
| \((0.07, 2.43)\) | \(2\) | \(y''(2)= -10\) | \(y\) is Concave Down |
| \((2.43, \infty)\) | \(3\) | \(y''(3)= 20\) | \(y\) is Concave Up |
Therefore, we can determine that the Points of Inflection are \(x = 0.66, 2.43\).
Given all the information we have gathered, we can now sketch our graph:
