Graphs & transformations

The diagram above shows the graph of $y=f(x)$.

The diagram shows the graph of $y=f(x)$. The curve passes through the origin $O$, the points $A(-4a,0)$ and $C(-2a,0)$. The points $B(-3a,2)$ and $D(-a,-2)$ are stationary points and the lines $y=0$ and $x=a$ are asymptotes of the curve.

The curve $C_1$ with equation $y=\dfrac{(x+2)^2}{x+1}$ is transformed onto the curve $C_2$ with equation $y=f(x)$. The curve $C_1$ has a minimum turning point $(0,4)$ which corresponds to the point with coordinates $(a,b)$ on the curve $C_2$, where $a,b>0$. Given that $f(x)$ has the form $\dfrac{p^2x^2}{px-1}+q$, where $p,q$ are positive constants, express…

The diagram below shows the graph of $y=f(x)$. The curve cuts the axes at $(-1,0)$ and $(0,3)$. It has a maximum point at $(3,5)$, and asymptotes $x=-2$ and $y=2$.

A curve has equation $y=f(x)$, where

(a) A curve $y=f(x)$ is transformed by a stretch with scale factor 2 parallel to the $y$-axis, followed by a reflection about the $x$-axis. The equation of the resultant curve is $y=\ln[\dfrac1{(x+a)^2}]$, where $x>-a$ and $a>1$. Find $f(x)$ and sketch the graph of $y=f(x)$.

(a) Describe a sequence of exactly three transformations that will transform the curve with equation $y=\dfrac1{x-3}$ onto the curve with equation $y=\dfrac{2x+a}{3-x}$, where $a$ is a positive constant.

The ellipse $C_1$ with the equation $\dfrac{x^2}{6^2}+\dfrac{y^2}{3^2}=1$ is transformed by a stretch with scale factor $k$ parallel to the $x$-axis, followed by a translation of $m$ units in the positive $x$-direction, followed by a translation of $n$ units in the negative $y$-direction, to form $C_2$.

The curve $C$ has equation $y=\dfrac{ax+b}{x^2+c}$, where $a,b,c$ are constants. It is given that $x=2$ is an asymptote of $C$.

Sketch the graphs of $y=\ln(1-|x|)$ and $y=-|x-1|$ on the same diagram, showing clearly the equations of any asymptotes and any intersections with the axes.

The curve $C$ has equation given by $y=\dfrac{x-a+a^2}{a(x-a)}$, where $a$ is a negative constant.

The curve $C$ has equation $y=\dfrac{x^2+b}{x+a}$, where $a>0,b>0$.

The diagram below shows a sketch of the graph of $y=\dfrac{ax^2+bx+c}{x+d}$ where $a,b,c,d$ are constants. The asymptotes, also shown in the diagram, are $x=-2$ and $y=x-4$.

The curve $C$ has equation $y=\dfrac{ax^2+x-4}{x+2}$, where $a$ is a constant.

The curve $C$ has equation $y=ax+b+2a+\dfrac{2(b+2a)}{x-2}$, where $a,b$ are real constants such that $a>0,b\ne-2a,x\ne2$.

The curves $C_1,C_2$ have equations $(x-2)^2=a^2(1-y^2)$ and $y=\dfrac{x^2-4}{x+1}$, where $1<a<2$, respectively. Describe the geometrical shape of $C_1$.

The curve $C_1$ has equation $\dfrac{y^2}{4}-x^2=1$ and the curve $C_2$ has equation $y=k-\dfrac1x$, where $k$ is a positive constant. The horizontal asymptote of $C_2$ is a tangent to $C_1$.

The curve $C$ has equation $y=\dfrac{2x^2+4}{3(x+\lambda+1)}$. It is given that $C$ has a vertical asymptote $x=0$.

The hyperbola $C$ passes through the point $(2,0)$ and has oblique asymptotes $y=-2x$ and $y=2x$.

A curve $C$ has parametric equation $x=t^2-1$, $y=4t-t^3+15$, where $t\in\mathbb R$.