JC1 H2 Mathematics: Graphing Techniques and Transformations

Sketch a rational curve completely

Sketch $y=(x^2+3)/(x-1)$, showing asymptotes, intercepts and stationary points.

The domain excludes $x=1$. As $x\to1^+$, $y\to+\infty$; as $x\to1^-$, $y\to-\infty$. Thus $x=1$ is a vertical asymptote. Since the final fraction tends to zero as $x\to\pm\infty$, the oblique asymptote is $y=x+1$.

At $x=0$, $y=-3$. The numerator $x^2+3$ is positive for every real $x$, so there are no x-intercepts.

At $x=-1$, $y=-2$. At $x=3$, $y=6$. The derivative is positive when $|x-1|>2$, and negative when $0<|x-1|<2$. Hence $(-1,-2)$ is a maximum and $(3,6)$ is a minimum. There is a break at the vertical asymptote between them.

The left branch lies below $y=x+1$ because $4/(x-1)<0$; the right branch lies above it. These checks fix the branches' positions.

Use a horizontal line and the discriminant

For the same curve, prove algebraically that $-2<y<6$ is impossible. Hence determine how many real roots $x^2+3=k(x-1)$ has as $k$ varies.

Intersect the curve with $y=k$ and rearrange as a quadratic in $x$:

Real intersections require $\Delta\geq0$, so $k\leq-2$ or $k\geq6$. The endpoints occur at the stationary points. Also $x=1$ cannot be a root of the rearranged equation: it would give $4=0$. Thus no invalid root has been introduced.

The range is $( -\infty,-2]\cup[6,\infty)$. The equation has two distinct real roots for $k<-2$ or $k>6$, one repeated real root for $k=-2$ or 6, and no real roots for $-2<k<6$.

Use the given asymptotes to recover coefficients

The curve $y=(ax^2+bx+c)/(x+d)$ has vertical asymptote $x=-1$ and oblique asymptote $y=2x-1$. Find $a,b,d$, then the condition for no stationary points.

The vertical asymptote gives $d=1$. Use the oblique asymptote as the quotient:

Comparing coefficients gives $a=2,b=1$ and $R=c+1$. The stated vertical asymptote requires $c\ne-1$.

For $c>-1$, there are two distinct permitted stationary x-values. For $c<-1$, the square would be negative, so there are none. Hence $c<-1$ is the required condition.

If $c=-1$, the rule simplifies to a straight line with a hole at $x=-1$. It fails the stated vertical-asymptote condition. A zero discriminant in an equation obtained after clearing denominators does not automatically give a stationary point.

A conic written in expanded form

Identify $4x^2+9y^2-8x+18y-23=0$, and give the equation and domain of its lower half.

This is an ellipse with centre $(1,-1)$, horizontal radius 3 and vertical radius 2. Its horizontal vertices are $(-2,-1),(4,-1)$; its vertical vertices are $(1,1),(1,-3)$.

For the lower half, $y+1\leq0$, so choose the negative square root:

The expression under the root must be non-negative. The endpoints are included and lie on the centre's horizontal line. Choosing the other sign gives the upper half.

Changing an ellipse can change four intersections into two

For $k>0$, find when the ellipse $k(x-1)^2+4(y+1)^2=4$ meets the hyperbola $(x-1)^2-(y+1)^2=1$ in exactly four points.

Put $U=(x-1)^2$ and $V=(y+1)^2$. Both must be non-negative.

Since $k>0$, $U>0$. Four distinct points require $V>0$, giving two signs for each coordinate offset. Hence $0<k<4$.

At $k=4$, $V=0$: only two points remain, $(0,-1)$ and $(2,-1)$. For $k>4$, $V<0$, so there are no real intersections.

For the illustrated value $k=1$, the offsets are $x-1=\pm\sqrt{8/5}$ and $y+1=\pm\sqrt{3/5}$. Substitution checks all four points. The asymptotes are $y+1=\pm(x-1)$.

Different parameter limits can give different vertical lines

A curve has $x=t^2-2t$, $y=1/(1-t^2)$, with $t<1$ and $t\ne-1$. Find its asymptotes.

The denominator vanishes at $t=\pm1$, but the asymptotes are lines in the x-y plane, not equations in $t$.

Thus $x=3$ is a vertical asymptote. At the excluded upper parameter endpoint,

Thus $x=-1$ is another vertical asymptote. Finally,

Hence $y=0$ is a horizontal asymptote. A finite limit of one coordinate gives an asymptote only when the other coordinate tends to infinity. Note that $x=(t-1)^2-1>-1$ on the given parameter interval.

An input change and an output change together

The graph $y=f(x)$ has minimum point $(1,-4)$ and x-intercepts $(-1,0),(3,0)$. Find their images on $y=-2f(3x+1)+5$, and describe a sequence of transformations.

A source point $(a,b)$ moves to an x-value satisfying $3x+1=a$, with new y-value $-2b+5$:

Thus the minimum becomes a maximum at $(0,13)$. The old x-intercepts move to $(-2/3,5)$ and $(2/3,5)$; they are not new x-intercepts.

One valid sequence is: translate 1 unit in the negative x-direction, scale by factor 1/3 parallel to the x-axis, scale by factor 2 parallel to the y-axis, reflect in the x-axis, then translate 5 units upwards.

The equation after each step checks the sequence. For $f(x)=(x-1)^2-4$, the final rule is $-18x^2+13$, which has maximum $(0,13)$.

Three derived graphs from the same quadratic

For $f(x)=x^2-2x=(x-1)^2-1$, sketch $|f(x)|$, $f(|x|)$, and $1/f(x)$.

The original roots are 0 and 2, and the minimum is $(1,-1)$. For $|f(x)|$, reflect only the part below the x-axis:

The reflected part has maximum $(1,1)$, and the roots remain 0 and 2. The roots become corners.

For $f(|x|)$, retain the original right half and reflect it to the left:

This even graph has minima $(-1,-1),(1,-1)$, and roots $-2,0,2$. The origin is a corner maximum, so it is not a stationary point.

For the reciprocal, the zeros of $f$ become vertical asymptotes:

The asymptotes are $x=0,x=2,y=0$. There are no axial intercepts. At $x=1$, $g=-1$, and the derivative changes from positive to negative: $(1,-1)$ is a maximum. The outer branches are positive; the central branch is negative.

Use stationary points and increasing or decreasing intervals

For $f(x)=x^3-3x$, sketch $f'$ and explain how its signs match the original curve.

The derivative graph is an upward parabola, with roots $(-1,0),(1,0)$, and minimum $(0,-3)$. It is positive outside $[-1,1]$ and negative inside. Thus $f$ increases up to its maximum $(-1,2)$, decreases to its minimum $(1,-2)$, then increases again.

At a maximum the derivative crosses from positive to negative; at a minimum it crosses from negative to positive. At a stationary point of inflexion it can touch zero without changing sign: for $f(x)=x^3$, $f^{\prime}(x)=3x^2$ touches zero at the origin.

The y-coordinate on a derivative graph is a gradient, not the original graph's height. An original intercept does not normally give a derivative intercept.

Choose the sign on each part

A differentiable function satisfies $|f(x)|=|x^2-1|$, with $f^{\prime}(x)<0$ for $x<0$ and $f^{\prime}(x)>0$ for $x>0$. Recover $f$.

The modulus gives the shape but loses the sign. For $x<-1$, the branch $x^2-1$ has negative gradient, so retain it above the axis. Between −1 and 0, the positive modulus branch $1-x^2$ has positive gradient; reflect it below the axis to get the required negative gradient.

Between 0 and 1, the reflected branch $x^2-1$ has positive gradient. For $x>1$, retain the positive outer branch, which also has positive gradient. All four pieces therefore join as

The function is differentiable at both roots and has minimum $(0,-1)$. Its modulus is the stated one, and $2x$ has exactly the required signs. Choosing every branch above the axis would fail the gradient information.