Read curve shape and derivative graphs

Concavity describes how the slope changes

Read the two derivatives separately: $f^{\prime}$ describes whether the function rises or falls; $f^{\prime\prime}$ describes whether its slope increases or decreases. A curve can fall while its slope increases.

The source defines the second derivative as $d(dy/dx)/dx=d^2y/dx^2$: the first derivative is the rate of change of y with respect to x; the second is the rate of change of that first derivative with respect to x.

If $f^{\prime\prime}(x)>0$ throughout an interval I, then the slope $f^{\prime}$ is increasing and f is concave upwards there. The page 35 illustrations show tangents below the curve, both for a decreasing curve on (a,b) and an increasing curve on (c,d).

If $f^{\prime\prime}(x)<0$ throughout I, the slope decreases and f is concave downwards: the tangents lie above the source’s increasing curve on (a,b) and decreasing curve on (c,d).

These are interval statements under the usual differentiability assumptions; they are not decided from a single tangent or from whether the function is above the x-axis.

Compare the four cases in the complete page 36 graphical summary:

Source case 1

• f′: positive.
• Function behaviour: Strictly increasing.
• f″: negative.
• Concavity: Downwards.

Source case 2

• f′: positive.
• Function behaviour: Strictly increasing.
• f″: positive.
• Concavity: Upwards.

Source case 3

• f′: negative.
• Function behaviour: Strictly decreasing.
• f″: negative.
• Concavity: Downwards.

Source case 4

• f′: negative.
• Function behaviour: Strictly decreasing.
• f″: positive.
• Concavity: Upwards.

The source asks whether strict increase implies $dy/dx>0$. The Teacher supplies No, with counterexample $y=x^3$: it is strictly increasing but has derivative zero at 0. This reinforces the earlier sufficient-test qualification, not a converse theorem. Example 4 preserves the source’s shuffled interval choices and Teacher matching arrows, while distinguishing rounded graph labels from exact derivative boundaries.

Inflection means a change of concavity

The source compares $y=x^3$ with $y=x^3+x$. Both change from concave down for $x<0$ to concave up for $x>0$ at $(0,0)$, and both have second derivative zero there.

For $x^3$, the first derivative is zero, so the inflection is stationary. For $x^3+x$, the first derivative is 1, so it is non-stationary; the source explicitly labels non-stationary inflection as outside its H2 syllabus scope. An inflection is not a local maximum or minimum in these configurations.

The source says an inflection has $f^{\prime\prime}=0$, but the converse fails. Its counterexample is $y=x^4$: $f^{\prime\prime}(0)=0$ while the curve stays concave upwards on both sides. The existence condition matters: if the second derivative exists at an inflection in this setting, it is zero.

Without that condition, $f(x)=x|x|$ changes concavity at 0 but has no second derivative there. This last example is an independently authored qualification, not a change to the retained source statement.

Sketch the function and derivative together

Page 38 asks for $Y_1=0.1X(X-6)(X+6)$ and its derivative using a G.C. The source callout shows $Y_2$ as the numerical derivative of $Y_1$ with respect to X, evaluated at the current X: use MATH → 8 for the derivative template and the illustrated ALPHA → TRACE → ENTER route to insert $Y_1$.

The printed graphs have original turning labels approximately $(-3.46,8.31)$ and $(3.46,-8.31)$, and derivative x-intercepts approximately $-3.46,3.46$. The Teacher fills the answer: turning points of f correspond to x-intercepts of f′.

Here the analytical derivative is $0.3x^2-3.6$, with exact zeros $\pm2\sqrt3$; a numerical derivative display is an approximation.

Both source QR codes point to the function-and-derivative exploration. This is a separately decoded source pointer; its external content has not been reviewed.

The source’s four-step derivative-sketch procedure

1. Inspect asymptotes. The printed table maps vertical $x=a$ to vertical $x=a$, horizontal $y=a$ to horizontal $y=0$, and oblique $y=ax+b$ to horizontal $y=a$ for the derivative graph.
2. Inspect stationary points. A maximum at (a,b) gives a derivative zero (a,0) crossed positive-to-negative; a minimum gives negative-to-positive. A decreasing stationary inflection gives a derivative local maximum touching(a,0) from below; an increasing stationary inflection gives a derivative local minimum touching(a,0) from above.
3. Inspect slope signs. Negative original gradient puts the derivative below the x-axis; positive original gradient puts it above.
4. Combine steps 1–3 into the sketch, preserving the original domain and every separate branch.

The source asymptote table is a sketching guide for the regular curves being considered, not a theorem for every differentiable function. Check the actual slope behaviour before transferring asymptotes.

For instance, on $x>0$, $f(x)=\sin(x^2)/x$ approaches horizontal asymptote 0, but $f^{\prime}(x)=2\cos(x^2)-\sin(x^2)/x^2$ does not approach 0. Adding $ax+b$ gives the same warning for an oblique asymptote; vertical asymptotes also require inspecting one-sided slopes. These qualifications are independently authored. The paired local sketches below preserve all four stationary-point correspondences.

Explain what the derivative graph cannot locate

Compare $f(x)=x^3-3x$ with $g(x)=x^3-3x+5$. Predict how their function graphs differ and how their derivative graphs compare. Which stationary-point information can the derivative graph determine, and what extra information is needed for the heights?

After your attempt: compare the reasoning

Both derivatives are $3x^2-3$. Both functions increase for $x<-1$ and $x>1$, and decrease for $-1<x<1$. Each has a local maximum at input $-1$ and a local minimum at input $1$.

For $f$, the corresponding heights are 2 and $-2$; for $g$, they are 7 and 3.

A vertical translation leaves slopes unchanged, so a derivative graph alone does not fix these heights. A value of the function at one input would fix the translation in this comparison.