Graphs and transformations – Year 12 A level maths revision notes to download as PDF

Worked example 1

Sketch \(y=(x-1)(x+2)(x-3)\).

Roots: \(1,-2,3\). The \(y\)-intercept is \((-1)(2)(-3)=6\). The leading term is \(x^3\), so the graph rises to the right.

Worked example 2

Show that \(x^3=\dfrac1x\) has exactly two real solutions.

Sketch \(y=x^3\) and \(y=\dfrac1x\): they cross at two points, one in each of the first and third quadrants. Algebra confirms \(x^4=1\), so \(x=\pm1\).

Worked example 3

The curve \(y=x^2\) is translated by \(\begin{pmatrix}2\\1\end{pmatrix}\). Find the new equation.

The new equation is \(y=(x-2)^2+1\) and the vertex moves from \((0,0)\) to \((2,1)\).

Worked example 4

The point \((4,6)\) lies on \(y=f(x)\). Find its image on \(y=f(2x)\), \(y=2f(x)\) and \(y=f(x-1)\).

On \(y=f(2x)\) the \(x\)-coordinate is halved: \((2,6)\). On \(y=2f(x)\) the \(y\)-coordinate doubles: \((4,12)\). On \(y=f(x-1)\) the point moves right 1: \((5,6)\).

Worked example 5

Describe how \(y=x^2-6x+11\) is obtained from \(y=x^2\).

\(x^2-6x+11=(x-3)^2+2\), so it is \(y=x^2\) translated by \(\begin{pmatrix}3\\2\end{pmatrix}\).

Worked example 6: Sketching a cubic with a repeated root

Sketch \(y=x^2(3-x)\) and find where \(y>0\).

The factor \(x^2\) gives a repeated root at \(x=0\) (the curve touches the axis), and \((3-x)\) gives a root at \(x=3\) (the curve crosses). The coefficient of \(x^3\) is \(-1\) (negative), so the curve starts high on the left and ends low on the right. Between the roots \(y>0\) for \(0\lt x<3\) and also for \(x<0\) because \(x^2>0\) and \(3-x>0\). So \(y>0\) when \(x<3\) (excluding the touching point \(x=0\) where \(y=0\)).

Worked example 7: Sketching a reciprocal curve with a translation

Sketch \(y=\dfrac{4}{x+1}-2\), showing the asymptotes and the intercepts.

Start from \(y=\dfrac4x\) (two branches in the first and third quadrants). Translate 1 unit left and 2 units down. The vertical asymptote is \(x=-1\) and the horizontal asymptote is \(y=-2\). When \(x=0\): \(y=4-2=2\), so the curve meets the \(y\)-axis at \((0,2)\). When \(y=0\): \(\dfrac4{x+1}=2\), \(x+1=2\), \(x=1\), so it meets the \(x\)-axis at \((1,0)\). The upper branch passes through both intercepts and the lower branch lies below \(y=-2\).

Worked example 8: Combined transformation and order

The point \((2,5)\) lies on \(y=f(x)\). Find its image on the curve \(y=3f(x-1)+2\).

Work from the inside outwards. The translation \(x\to x-1\) moves the point 1 to the right: \((3,5)\). The stretch factor 3 parallel to the \(y\)-axis multiplies the \(y\)-coordinate: \((3,15)\). Finally add 2 to the \(y\)-coordinate: \((3,17)\). The order matters: if you added 2 first and then multiplied by 3 you would get 21, which is the image on \(y=3[f(x-1)+2]\), a different curve.

Worked example 9: f(ax) compared with af(x)

\(f(x)=x^2-1\). Find and compare \(f(2x)\) and \(2f(x)\).

\(f(2x)=(2x)^2-1=4x^2-1\). \(2f(x)=2(x^2-1)=2x^2-2\). They are different curves. For \(f(2x)\) the roots are at \(x=\pm\tfrac12\) (half as far from the axis as before) and the \(y\)-intercept is still \(-1\). For \(2f(x)\) the roots remain at \(x=\pm1\) and the \(y\)-intercept is \(-2\). So a stretch parallel to the \(x\)-axis changes the \(x\)-coordinates, while a stretch parallel to the \(y\)-axis changes the \(y\)-coordinates.

Worked example 10: Solving an equation using graphs

Show that the equation \(x^3-3x+1=0\) has three real roots and find an interval for each root.

Let \(f(x)=x^3-3x+1\). Then \(f(-2)=-1\), \(f(-1)=3\), \(f(0)=1\), \(f(1)=-1\), \(f(2)=3\). The sign changes between \(-2\) and \(-1\), between \(0\) and \(1\) and between \(1\) and \(2\), so there is a root in each interval and, since a cubic has at most three roots, exactly three real roots. More precisely they are approximately \(-1.88\), \(0.35\) and \(1.53\). Drawing \(y=x^3\) and the line \(y=3x-1\) on the same axes shows the same three intersections.

Worked example 11: Transforming an exponential graph

Sketch \(y=2^{x-1}+3\). State the asymptote and the \(y\)-intercept, and find where the curve crosses the line \(y=5\).

Start with \(y=2^x\), which passes through \((0,1)\) and has the \(x\)-axis as asymptote. Translate by \(\begin{pmatrix}1\3\end{pmatrix}\): the asymptote becomes \(y=3\) and the point \((0,1)\) moves to \((1,4)\). The \(y\)-intercept is \(2^{-1}+3=3.5\). The curve is always above the line \(y=3\). To meet \(y=5\) solve \(2^{x-1}=2\), giving \(x-1=1\) and \(x=2\). So the curve crosses \(y=5\) at the single point \((2,5)\).

Worked example 12: Equation of an image under a stretch and a translation

The curve \(y=x^3\) is stretched by scale factor 2 parallel to the \(y\)-axis and then translated by \(\begin{pmatrix}-1\4\end{pmatrix}\). Find the equation of the image and its \(y\)-intercept.

After the stretch the equation is \(y=2x^3\). Replacing \(x\) by \(x+1\) moves the curve one unit left, and adding 4 moves it up: \(y=2(x+1)^3+4\). When \(x=0\): \(y=2+4=6\). The point of inflection of \(y=x^3\) at the origin has moved to \((-1,4)\), which you can read directly from the equation. Expanding gives \(y=2x^3+6x^2+6x+6\).