Welcome to one of the most important synthesis topics in Calculus I. Until now, we've used derivatives to find individual pieces of information about a function: the slope of a tangent line, the velocity of a particle, or a single local maximum.
Now, we are going to put all of those pieces together. We will use limits, continuity, $f'(x)$, and $f''(x)$ to build a complete "blueprint" of a function's behavior.
The Goal: To be able to draw a detailed, accurate graph of a function $f(x)$ using only the tools of calculus, without relying on a graphing calculator.
The key to this process is organization. We will follow a consistent 8-step checklist for every function we analyze.
Find the set of all $x$-values for which $f(x)$ is defined. (Look for denominators, square roots, logarithms, etc.)
$y$-intercept: Plug in $x=0$ to find $f(0)$.
$x$-intercept(s): Set $f(x)=0$ and solve for $x$. (This is sometimes too hard, and we can skip it).
Find $f'(x)$, identify all critical points (where $f'=0$ or $f'$ DNE), and build a sign chart for $f'(x)$.
Use the sign chart from Step E.
A local max occurs where $f'$ switches from $+$ to $-$.
A local min occurs where $f'$ switches from $-$ to $+$.
Find the $y$-values for each of these points.
Find $f''(x)$, identify points where $f''=0$ or $f''$ DNE, and build a sign chart for $f''(x)$.
Find the $(x,y)$ coordinates of any inflection points (where concavity changes).
1. Draw and label all asymptotes (as dashed lines).
2. Plot all key points (intercepts, local max/mins, inflection points).
3. Use the $f'$ and $f''$ charts to connect the dots, ensuring the curve rises/falls and bends in the correct way in each interval.
$y$-intercept: $f(0) = 0 - 0 - 0 + 1 = 1$. The point is $(0, 1)$.
$x$-intercept(s): We would need to solve $2x^3 - 3x^2 - 12x + 1 = 0$. This is difficult, so we will skip this step and let our final sketch show us roughly where they are.
$f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x-2)(x+1)$
Critical points are at $x=2$ and $x=-1$.
Sign chart for $f'$:
Interval $(-\infty, -1)$: Test $x=-2 \implies 6(-)(-)=+$. Increasing.
Interval $(-1, 2)$: Test $x=0 \implies 6(-)(+)=-$. Decreasing.
Interval $(2, \infty)$: Test $x=3 \implies 6(+)(+)=+$. Increasing.
At $x=-1$, $f'$ switches from $+$ to $-$. This is a local max.
$f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 1 = -2 - 3 + 12 + 1 = 8$.
Max point: $(-1, 8)$.
At $x=2$, $f'$ switches from $-$ to $+$. This is a local min.
$f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 1 = 16 - 12 - 24 + 1 = -19$.
Min point: $(2, -19)$.
$f''(x) = 12x - 6 = 6(2x - 1)$
Potential inflection point at $x=1/2$.
Sign chart for $f''$:
Interval $(-\infty, 1/2)$: Test $x=0 \implies 6(-)=-$. Concave Down.
Interval $(1/2, \infty)$: Test $x=1 \implies 6(+)=+$. Concave Up.
Concavity changes at $x=1/2$, so it is an inflection point.
$f(1/2) = 2(\frac{1}{8}) - 3(\frac{1}{4}) - 12(\frac{1}{2}) + 1 = \frac{1}{4} - \frac{3}{4} - 6 + 1 = -\frac{1}{2} - 5 = -5.5 = -11/2$.
Inflection point: $(1/2, -11/2)$.
We collect our key points:
$y$-intercept: $(0, 1)$
Local Max: $(-1, 8)$
Local Min: $(2, -19)$
Inflection Point: $(0.5, -5.5)$
We draw a curve that passes through these points, making sure it is increasing and concave down on $(-\infty, -1)$, decreasing and concave down on $(-1, 1/2)$, decreasing and concave up on $(1/2, 2)$, and increasing and concave up on $(2, \infty)$.
$y$-intercept: $f(0) = \frac{0^2}{0^2 - 4} = 0$. The point is $(0, 0)$.
$x$-intercept(s): Set $f(x)=0 \implies \frac{x^2}{x^2 - 4} = 0 \implies x^2 = 0 \implies x=0$.
The only intercept is the origin $(0, 0)$.
$f(-x) = \frac{(-x)^2}{(-x)^2 - 4} = \frac{x^2}{x^2 - 4} = f(x)$.
The function is Even. It is symmetric with respect to the $y$-axis.
Vertical (VA): The denominator is zero at $x=2$ and $x=-2$. The numerator is not zero at these points, so we have vertical asymptotes.
VA at $x=2$ and $x=-2$.
Horizontal (HA): We check the limit as $x \to \infty$.
$\lim_{x\to\infty} \frac{x^2}{x^2 - 4} = \lim_{x\to\infty} \frac{x^2/x^2}{(x^2-4)/x^2} = \lim_{x\to\infty} \frac{1}{1 - 4/x^2} = \frac{1}{1-0} = 1$.
HA at $y=1$. (The limit is the same for $x \to -\infty$ since the function is even).
Slant (SA): The degree of the numerator (2) is not one more than the denominator (2), so there is no slant asymptote.
$f'(x) = \frac{(2x)(x^2-4) - (x^2)(2x)}{(x^2-4)^2} = \frac{2x^3 - 8x - 2x^3}{(x^2-4)^2} = \frac{-8x}{(x^2-4)^2}$
Critical point (where $f'=0$) is at $x=0$.
Points where $f'$ DNE are $x=\pm 2$ (already known as VAs).
Sign chart for $f'$: (Note: denominator $(x^2-4)^2$ is always positive)
Interval $(-\infty, -2)$: Test $x=-3 \implies \frac{-8(-3)}{+} = +$. Increasing.
Interval $(-2, 0)$: Test $x=-1 \implies \frac{-8(-1)}{+} = +$. Increasing.
Interval $(0, 2)$: Test $x=1 \implies \frac{-8(1)}{+} = -$. Decreasing.
Interval $(2, \infty)$: Test $x=3 \implies \frac{-8(3)}{+} = -$. Decreasing.
At $x=0$, $f'$ switches from $+$ to $-$. This is a local max.
We already know this point: $(0, 0)$.
$f''(x) = \frac{(-8)(x^2-4)^2 - (-8x)[2(x^2-4)(2x)]}{(x^2-4)^4}$
...divide out one $(x^2-4)$...
$f''(x) = \frac{-8(x^2-4) - (-8x)(4x)}{(x^2-4)^3} = \frac{-8x^2 + 32 + 32x^2}{(x^2-4)^3} = \frac{24x^2 + 32}{(x^2-4)^3}$
The numerator $24x^2 + 32$ is always positive. The sign of $f''$ depends only on the sign of the denominator $(x^2-4)^3$.
Sign chart for $f''$:
Interval $(-\infty, -2)$: Test $x=-3 \implies \frac{+}{(-3)^2-4)^3} = \frac{+}{(+)^3}=+$. Concave Up.
Interval $(-2, 2)$: Test $x=0 \implies \frac{+}{(0-4)^3} = \frac{+}{(-)^3}=-$. Concave Down.
Interval $(2, \infty)$: Test $x=3 \implies \frac{+}{(3^2-4)^3} = \frac{+}{(+)^3}=+$. Concave Up.
No inflection points, as concavity only changes at the VAs.
We draw VAs at $x=\pm 2$ and a HA at $y=1$.
We plot the single point $(0,0)$, which is a local max.
On $(2, \infty)$, the curve is Dec and CU, "squeezed" between $y=1$ and $x=2$.
On $(0, 2)$, the curve starts at $(0,0)$, is Dec and CD, and heads to $-\infty$ at $x=2$.
By symmetry, we mirror this on the other side.
Let's apply the 8-step guideline to a function that has a Slant Asymptote.
$y$-intercept: $f(0) = \frac{0^2}{0-1} = 0$. Point: $(0, 0)$.
$x$-intercept(s): Set $f(x)=0 \implies x^2 = 0 \implies x=0$.
The only intercept is the origin $(0, 0)$.
$f(-x) = \frac{(-x)^2}{(-x)-1} = \frac{x^2}{-x-1}$. This is not $f(x)$ or $-f(x)$. No symmetry.
Vertical (VA): Denominator is zero at $x=1$. $\boxed{\text{VA at } x=1}$.
Horizontal (HA): $\lim_{x\to\infty} \frac{x^2}{x-1} \to \infty$. No HA.
Slant (SA): The degree of the numerator (2) is exactly one more than the denominator (1). A slant asymptote exists. We must use polynomial long division.
x + 1
_______
x-1 | x^2 + 0x + 0
-(x^2 - 1x)
---------
1x + 0
-(1x - 1)
-------
1
This shows $f(x) = (x+1) + \frac{1}{x-1}$. As $x \to \infty$, the $\frac{1}{x-1}$ term goes to 0, so the function $f(x)$ approaches the line $x+1$.
$f'(x) = \frac{(2x)(x-1) - (x^2)(1)}{(x-1)^2} = \frac{2x^2-2x-x^2}{(x-1)^2} = \frac{x^2-2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}$
Critical points (where $f'=0$) are at $x=0$ and $x=2$.
Point where $f'$ DNE is $x=1$ (our VA).
Sign chart for $f'$:
Interval $(-\infty, 0)$: Test $x=-1 \implies \frac{(-)(-)}{(+)} = +$. Increasing.
Interval $(0, 1)$: Test $x=0.5 \implies \frac{(+)(-)}{(+)} = -$. Decreasing.
Interval $(1, 2)$: Test $x=1.5 \implies \frac{(+)(-)}{(+)} = -$. Decreasing.
Interval $(2, \infty)$: Test $x=3 \implies \frac{(+)(+)}{(+)} = +$. Increasing.
At $x=0$, $f'$ switches from $+$ to $-$. This is a local max.
$f(0) = 0$. Point: $(0, 0)$.
At $x=2$, $f'$ switches from $-$ to $+$. This is a local min.
$f(2) = \frac{2^2}{2-1} = 4$. Point: $(2, 4)$.
From $f'(x) = \frac{x^2-2x}{(x-1)^2}$, we use the quotient rule:
$f''(x) = \frac{(2x-2)(x-1)^2 - (x^2-2x)[2(x-1)]}{(x-1)^4}$
...divide out $(x-1)$...
$f''(x) = \frac{(2x-2)(x-1) - 2(x^2-2x)}{(x-1)^3} = \frac{2(x-1)^2 - 2x^2+4x}{(x-1)^3}$
$f''(x) = \frac{2(x^2-2x+1) - 2x^2+4x}{(x-1)^3} = \frac{2x^2-4x+2 - 2x^2+4x}{(x-1)^3}$
$f''(x) = \frac{2}{(x-1)^3}$
The numerator is always positive. The sign depends only on $(x-1)^3$.
Sign chart for $f''$:
Interval $(-\infty, 1)$: Test $x=0 \implies \frac{+}{(-)^3} = -$. Concave Down.
Interval $(1, \infty)$: Test $x=2 \implies \frac{+}{(+)^3} = +$. Concave Up.
No inflection points, as concavity only changes at the VA.
We draw the VA at $x=1$ and the SA at $y=x+1$.
We plot the local max $(0,0)$ and local min $(2,4)$.
On $(-\infty, 0)$, the curve is Inc and CD, approaching the SA.
On $(0, 1)$, the curve is Dec and CD, starting from $(0,0)$ and diving down to $-\infty$ along the VA.
On $(1, 2)$, the curve is Dec and CU, starting from $+\infty$ along the VA and turning at the local min $(2,4)$.
On $(2, \infty)$, the curve is Inc and CU, heading up towards the SA.
Problem: Fully sketch the function $f(x) = \frac{x^2 - 3x + 2}{x^2 + 3x + 2}$ by following the 8-step guideline.
$y$-intercept: $f(0) = \frac{0-0+2}{0+0+2} = 1$. Point: $(0, 1)$.
$x$-intercept(s): Set numerator to 0. $(x-1)(x-2)=0$.
Points: $(1, 0)$ and $(2, 0)$.
$f(-x) = \frac{(-x)^2 - 3(-x) + 2}{(-x)^2 + 3(-x) + 2} = \frac{x^2 + 3x + 2}{x^2 - 3x + 2}$.
This is not $f(x)$ or $-f(x)$. No symmetry.
Vertical (VA): From the domain, we have VAs at $\boxed{x=-1 \text{ and } x=-2}$.
Horizontal (HA): $\lim_{x\to\infty} \frac{x^2 - 3x + 2}{x^2 + 3x + 2} = \lim_{x\to\infty} \frac{1 - 3/x + 2/x^2}{1 + 3/x + 2/x^2} = \frac{1}{1} = 1$.
$\boxed{\text{HA at } y=1}$.
Slant (SA): Degrees are equal. No SA.
$f'(x) = \frac{(2x-3)(x^2+3x+2) - (x^2-3x+2)(2x+3)}{(x^2+3x+2)^2}$
Numerator: $(2x^3 + 3x^2 - 5x - 6) - (2x^3 - 3x^2 - 5x + 6) = 6x^2 - 12 = 6(x^2 - 2)$.
$f'(x) = \frac{6(x^2 - 2)}{(x^2+3x+2)^2}$. CPs at $x=\pm\sqrt{2}$. (approx $\pm 1.414$).
Sign chart for $f'$: Denominator is always $(+)$.
$(-\infty, -2)$: Test $x=-3 \implies \frac{6(9-2)}{+} = +$. Increasing.
$(-2, -\sqrt{2})$: Test $x=-1.5 \implies \frac{6(2.25-2)}{+} = +$. Increasing.
$(-\sqrt{2}, -1)$: Test $x=-1.2 \implies \frac{6(1.44-2)}{+} = -$. Decreasing.
$(-1, \sqrt{2})$: Test $x=0 \implies \frac{6(-2)}{+} = -$. Decreasing.
$(\sqrt{2}, \infty)$: Test $x=2 \implies \frac{6(4-2)}{+} = +$. Increasing.
At $x=-\sqrt{2}$, $f'$ switches from $+$ to $-$. Local Max.
$f(-\sqrt{2}) = \frac{(-\sqrt{2})^2 - 3(-\sqrt{2}) + 2}{(-\sqrt{2})^2 + 3(-\sqrt{2}) + 2} = \frac{2+3\sqrt{2}+2}{2-3\sqrt{2}+2} = \frac{4+3\sqrt{2}}{4-3\sqrt{2}}$
$= \frac{4+3\sqrt{2}}{4-3\sqrt{2}} \cdot \frac{4+3\sqrt{2}}{4+3\sqrt{2}} = \frac{16 + 24\sqrt{2} + 18}{16-18} = \frac{34+24\sqrt{2}}{-2} = \boxed{-17 - 12\sqrt{2} \approx -33.97}$.
At $x=\sqrt{2}$, $f'$ switches from $-$ to $+$. Local Min.
$f(\sqrt{2}) = \frac{(\sqrt{2})^2 - 3(\sqrt{2}) + 2}{(\sqrt{2})^2 + 3(\sqrt{2}) + 2} = \frac{2-3\sqrt{2}+2}{2+3\sqrt{2}+2} = \frac{4-3\sqrt{2}}{4+3\sqrt{2}}$
$= \frac{4-3\sqrt{2}}{4+3\sqrt{2}} \cdot \frac{4-3\sqrt{2}}{4-3\sqrt{2}} = \frac{16 - 24\sqrt{2} + 18}{16-18} = \frac{34-24\sqrt{2}}{-2} = \boxed{-17 + 12\sqrt{2} \approx -0.03}$.
(Work is very long) $f''(x) = \frac{-12(x^3 - 6x - 6)}{(x^2+3x+2)^3}$.
The cubic $g(x)=x^3 - 6x - 6$ has one real root $x_0 \approx 2.85$.
The sign of $f''$ depends on the signs of $-12(x-x_0)$ and $(x+1)^3(x+2)^3$.
$(-\infty, -2)$: $\frac{-(-)}{(-)^3(-)^3} = \frac{+}{+} = +$. Concave Up.
$(-2, -1)$: $\frac{-(-)}{(-)^3(+)^3} = \frac{+}{-} = -$. Concave Down.
$(-1, x_0)$: $\frac{-(-)}{(+)^3(+)^3} = \frac{+}{+} = +$. Concave Up.
$(x_0, \infty)$: $\frac{-(+)}{(+)^3(+)^3} = \frac{-}{+} = -$. Concave Down.
Concavity changes at VAs $x=-2, x=-1$ and at $x=x_0$.
Inflection Point at $x_0 \approx 2.85$.
We draw VAs at $x=-2, x=-1$ and HA at $y=1$. We plot intercepts $(0,1), (1,0), (2,0)$ and the local max $(-\sqrt{2}, -33.97)$ and local min $(\sqrt{2}, -0.03)$.
We draw the curve respecting all Inc/Dec and Concavity intervals.