Graphs of functions | Oxford Maths Admissions Test Livestream

Graphs Solutions

Part of the Oxford Maths Admissions Test Livestream 2026

These are the solutions for the Graphs worksheet

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Revision Questions

  1. Note that $x^2+4x+3=(x+3)(x+1)$. The graph of $y=f(x+2)$ is the graph of $y=f(x)$ after it has been translated two units to the left. 
    $f(x)$
    A quadratic graph with two roots, both values negative.
    $f(x+2)$
    A quadratic graph with two roots, both values even more negative than before.
    For $y=3f(2x)$, the graph is "squashed" by a factor of 2 parallel to the $x$-axis, then "stretched" by a factor of 3 parallel to the $y$-axis. 
    A quadratic graph with two negative roots, now both quite close to the origin, with the parabola narrower (or sharper) than before.
    For $y=2 f( 3 x)$, the graph is squashed by a factor of 3 parallel to the $x$-axis, then stretched by a factor of 2 parallel to the $y$-axis 
    Another quadratic graph. This one is just as narrow/sharp as the previous one, but the roots are in slightly different places.
    It's not the same as the previous graph. For example, the roots aren't in the same places. 
    $g(x)=x$ works for the last part; then $y=20x$ in both cases.
  2. The graph of $y=f(x)$ and the graph of $y=2f(x+1)$; 
    $f(x)=x^3-x$
    A classic cubic shape; the graph increases, then decreases, then increases, with two turning points along the way. Beautiful.
    $2f(x+1)$
    The same classic cubic shape, but it's translated to the left a bit, and the values are slightly larger than before.
    The graph of $y=2f(x)+1$; 
    $2f(x)+1$
    The same cubic shape, but it's been translated upwards this time. This means that it only has one real root instead of the three that it had before.
    It's not the same as the previous graph. We could compare, for example, the values of $2f(x+1)$ and $2f(x)+1$ when $x=0$. 
    $g(x)=x/3+c$ for any constant $c$ works for the last part.
  3. For large $n$, $y=x^n$ is close to zero between $-1$ and $1$, except for values near $x=\pm 1$. For small positive $n<1$, $y=x^n$ is close to $1$ between $0$ and $1$, except for values near $x=0$. For negative $n$, the graph increases without bound near $x=0$. 
    $y=x^9$
    A graph that's very flat close to zero between -1 and 1, except near those points; it's suddenly very negative near x = -1, and it's suddenly very positive near x = 1.
    $y=x^{1/5}$
    A graph that grows astonishingly quickly near x = 0 but then slows down, growing slower and slower after that.
    $y=x^{-1/2}$
    A graph that's decreasing. Near x = 0, the values are extremely large, and the graph disappears off the top of the axes. For larger values of x, the graph tends towards the x-axis.
  4. Note that $\sqrt{4x+1}=2\sqrt{x+\frac{1}{4}}$ so this is a translation of the graph of $y=\sqrt{x}$ by $\frac{1}{4}$ units in the negative $x$-direction followed by a stretch parallel to the $y$-axis with scale factor 2. 
    $y=\sqrt{4x+1}$
    A square-root graph that's been translated left and stretched.
  5. If $x\geq0$ then $\sqrt{x^2}=x$ but if $x<0$ then $\sqrt{x^2}=-x$, because $\sqrt{u}$ is always the non-negative root. 
    $y=\sqrt{x^2}$
    The graph of the modulus function.
  6. The function $\sin(x^2)=0$ when $x^2= 180^\circ n$ for $n$ an integer, so the graph crosses the $x$-axis more and more frequently as $x$ increases. The graph has reflectional symmetry in the $y$-axis. 
    $y=\sin(x^2)$
    A distorted sine graph; the further we get from the origin the faster the curve oscillates. Near the origin itself, the graph looks like a parabola, passing through the origin with zero derivative. The graph has reflectional symmetry in the y axis.
  7. Note that $\log_2(x^2-2x+1)=\log_2((x-1)^2)$. For $x>1$ this is just $2\log_2(x-1)$. For $x<1$, it is $2\log_2 (-(x-1))$, because for negative $x$, $\sqrt{x^2}=-x$. 
    $y=\log_2 x$
    A standard log graph
    $y=\log_2 (x^2-2x+1)$
    A standard log graph has been translated to the right by one unit, and its mirror image in the line x=1 has been added to the plot, to give the picture reflectional symmetry in the line x=1.
  8. The graph of $y=2^x$, and the graph of $y=2^{-x}$. 
    $y=2^x$
    A standard increasing exponential graph
    $y=2^{-x}$
    The mirror image of the previous graph; a standard decreasing exponential graph, made by reflecting the previous graph in the y-axis.
    Each graph is the reflection in the $y$-axis of the other graph.
  9. The graph of $y=\frac{1}{2} + \frac{1}{2}\cos 2x$ is related to the graph of $y=\cos x$ by a stretch parallel to the $x$-axis with scale factor $\frac{1}{2}$, a stretch parallel to the $y$-axis with scale factor $\frac{1}{2}$, and then a translation parallel to the $y$-axis by $\frac{1}{2}$ a unit. After those transformations, the local minima of $\cos x$ will be transformed to be on the $x$-axis. 
    $\cos x$
    A standard cosine graph.
    $\frac{1}{2}+\frac{1}{2}\cos 2x$
    A cosine graph that makes two oscillations instead of one, as we go from 0 to 360 degrees. The values have been increased so that the minima sit on the x-axis. The maxima still have y = 1.
  10. The equation $y=4-x$ is the equation of a straight line. 
    The equation $y=4-x^2$ is the equation of a parabola. 
    The equation $y^2=4-x^2$ is the equation of a circle with radius 2 centred on the origin. 
    $y=4-x$
    A straight line through the points (0,4) and (4,0).
    $y=4-x^2$
    A standard parabola that's pointing down. It passes through the points (0,4) and (2,0) and (-2,0).
    $y^2=4-x^2$
    A circle of radius 2 centered on the origin.
  11. If $\cos x = \cos y$ then either $x=y+360^\circ n$ for some integer $n$, or $x=-y+360^\circ n$ for some integer $n$. These are the equations of straight lines. 
    A square grid of straight lines at 45 degrees to the axes.
  12. It's helpful to consider different values of the function separately. Let's suppose that $f(x)=c$ and $f(y)=c$ for some particular value of $c$. 
    If $c$ is large enough then there is just one input to $f$ that gives the output $c$, so we must have $x=y$. In fact, it's always possible that $x=y$, so that line is part of our graph. 
    For some values of $c$ though, there are three possible inputs to $f$ that give that output. As we change $c$, these alternative solutions appear at the turning point of $y=f(x)$. In terms of $x$ and $y$, these solutions trace out a neat round graph. That part of the graph is actually an ellipse, but you are not expected to spot that! 
    Algebraically, if $x^3-x=y^3-y$ then we can rearrange for $x^3-y^3-x+y=0$ and factorise for $(x-y)(x^2+xy+y^2-1)$. You are not expected to know that $x^2+xy+y^2=1$ is the equation of an ellipse. 
    $y=f(x)$
    A standard cubic graph, with a horizontal line drawn across the graph with y-value jus less than the local maximum of the cubic.
    $f(x)=f(y)$
    A straight line y=x and an ellipse; this is like a circle that's been stretched in a direction that's at right angles to the straight line y=x.
  13. The equation $x^4+2x^2y^2+y^4-3x^2-3y^2+2=0$ simplifies to $(x^2+y^2)^2-3(x^2+y^2)+2=0$, which is a quadratic for $x^2+y^2$, with roots $x^2+y^2=1$ or $x^2+y^2=2$. This is a pair of circles. 
    Two concentric circles.
  14. The equation $x^6+3x^4y^2+3x^2y^4+y^6=1$ simplifies to $(x^2+y^2)^3=1$ so we have $x^2+y^2=1$ and this is a circle. 
    One circle.
  15. The equation rearranges to $x^3-xy=x^2y^2-y^3$. Take out a factor of $x$ on the left and $y^2$ on the right to factorise this as $x(x^2-y)=y^2(x^2-y)$. So either $x^2=y$ or $x=y^2$. This is a pair of parabolas. 
    Two parabolas; one standard upward-pointing parabola, and one that's been rotated by 90 degrees clockwise.

 

TMUA Questions

TMUA 2020 Paper 2 Question 5

  • The function involves $2^x$ twice, so I'd like to think about what happens to $2^x$ for different values of $x$.
  • If $x$ is very large and positive, $2^x$ is very large indeed. What does that tell me about the graph? Informally, I'll have $y=\dfrac{\text{large}}{1+\text{large}}$. That's about 1.
  • For very negative $x$, $2^x$ is about 0. What does that tell me about the graph? I'll have $y\approx 0$.
  • When $x=0$, I get $\displaystyle \frac{1}{1+1}=\frac{1}{2}$. Halfway in between.
  • I also sketched $\displaystyle y=\frac{u}{1+u}$, just to check my understanding. This is an increasing function of $u$ for $u>-1$. 
    An increasing function that passes through the origin. The axes are labelled u (horizontal) and y (vertical). The derivative decreases with u.
  • Asymptotes are not mentioned on the TMUA Content Specification, but the idea is that, if $f(x)$ approaches a value for large $|x|$ (either very positive or very negative), then we can draw a horizontal dotted line with that value, if we want to. You've maybe met this idea for graphs like $\displaystyle y=\frac{1}{x}$ which we say has a horizontal asymptote at $y=0$, thinking about the behaviour of $y$ for very large $x$.
  • The graph in this question has a horizontal asymptote at $y=1$, thinking about the values when $x$ is very large, and it also has a horizontal asymptote at $y=0$, thinking about the values when $x$ is very negative.
  • The answer is A.

 

Extension

  • Let $\displaystyle \mathrm{f}(x)= \frac{2^x}{1+2^x}$. Prove that $\mathrm{f}(-x)=1-\mathrm{f}(x)$. 
    Interpret this as a symmetry of the graph of $y=\mathrm{f}(x)$.
  • True or false? "If $\mathrm{f}(x)$ is increasing and $\mathrm{g}(x)$ is increasing, then $\mathrm{f}(\mathrm{g}(x))$ is increasing."
  • True or false? "If $\mathrm{f}(x)$ is decreasing and $\mathrm{g}(x)$ is decreasing, then $\mathrm{f}(\mathrm{g}(x))$ is decreasing."

 

TMUA 2021 Paper 2 Question 13

  • I sketched $y-x=3$ and $y-x^2=1$ on the same axes. 
    An upwards-pointing quadratic with no real roots, with reflectional symmetry in the y-axis, and a straight line with positive gradient that crosses the parabola in two places.
  • I had to think hard about the inequalities. Which region do I want? The region that's "below" both curves, I think, because I can re-write each of them as $y<[\text{something}]$.
  • It's strange to me that $y$ can go very negative; when I drew the curves I was expecting $R$ to be the finite region bounded between the curves (possibly because I'm used to integration questions that ask me to find the area of such a region).
  • The first inequality asks us whether the values of $x$ are limited for every point in $R$. No, $x$ can be very negative. At this point I went back to look at the inequalities very carefully, because this is precisely the sort of inequality that I'd get if $R$ were the finite region between the curves. It's not though.
  • II looks like a trap to me; I can see why you might think that this is true, by multiplying the inequalities together. But we know that this can go wrong, if both brackets are negative. Can I actually do that for the region $R$? Yes if I move away from the curves far enough, I think. Looking for a counter-example, I'll set $x=0$ and then I'm looking for $y$ such that $y<3$ and $y<1$ (to obey the given inequalities) and also $y^2 \geq 3$ (to contradict II). I'll set $y=-2$. One counter-example $(0,-2)$ is enough to disprove II.
  • III asks whether the values of $y$ are limited. No, $y$ can be large. Again, this is the sort of inequality that I might get if $R$ were the finite region between the curves. It's still not though.
  • None of them are true.
  • The answer is A.

 

Extension

  • Change $y-x^2<1$ to $y-x^2>1$. Which of I, II, III is/are true for every point in $R$?
  • If I were using this as the start of an interview question, then I might build towards a sketch of graph of $(y-x)(y-x^2)=3$. We might start by exploring how many points are on that graph for each value of $x$. For large $x$, what are the approximate values of $y$ (in terms of $x$) for those points?

 

TMUA 2021 Paper 1 Question 17

  • I'll want $x^2+y^2$ to be equal to $\displaystyle \frac{\pi}{6}$ or $\displaystyle \frac{5\pi}{6}$, or one of those plus $2\pi$ or $4\pi$, or so on.
  • Those are concentric circles (the centre is the origin) with different radii.
  • Since $x^2+y^2=r^2$ for a circle, the radius is the square root of each of the numbers I listed above.
  • These solutions for $r^2$ come in pairs, with a gap of $\displaystyle \frac{4\pi}{6}$ between the two solutions in the pair.
  • Thinking about the radius $r$ instead of $r^2$, I would like to take the square root of those numbers. What will happen to the gap between a pair of solutions?
  • Because the graph of $y=\sqrt{x}$ grows at a slower rate for larger values of $x$, if we look at a larger pair of solutions for $r^2$, then the gap between the values of $r$ will be smaller than it would have been for a smaller pair. 
    A square root graph and several straight lines. The horizontal axis is labelled r-squared, and there are two pairs of vertical lines from that axis up to the graph. Each pair of vertical lines has the same spacing, but the pairs of horizontal lines do not; for larger values of r-squared, the difference between values of r is smaller.
  • So I'm expecting pairs of circles, with a gap between the pair that decreases the further from the origin I look.
  • Options B and E have equally-spaced pairs.
  • Option C doesn't seem to have pairs, or any change in the gaps.
  • I have no idea what's going on with option D, but I don't like it.
  • Option A seems to fit the description above.
  • The answer is A.

 

Extension

  • Check your understanding; sketch the solutions to $\cos\left(x^2+y^2\right)=\frac{1}{2}$.
  • If I were using this as the start of an interview question, then next I might switch to a different function entirely, and ask you about the graph \[\left(x^2+y^2\right)^{1/2}\sin\left(x^2+y^2\right)=1.\]

 

TMUA 2021 Paper 1 Question 18

  • I drew a graph. This is a "sideways" parabola where the roles of $x$ and $y$ have been swapped compared with the usual parabolas I've seen before. For some values of $x$ there are no points on the graph, for some values there are two solutions for $y$ from the quadratic formula, and for one value of $x$ there's a single value of $y$ on the graph. 
    A sideways parabola with a turning point in the first quadrant. The point P in the second quadrant is labelled.
  • I can complete the square to find $x=(y-3)^2+2$, so the turning point (if we can call it that, the parabola is sideways!) is at $(2,3)$.
  • Nice, that turning point is on the same horizontal line as the point $P$.
  • So after rotation the turning point will be directly below $P$, a distance of 4 away, the same distance as before the rotation.
  • I drew a sketch of this. I've drawn a parabola for the new curve (partly because all the options are parabolas), and I've drawn it with a negative leading coefficient (pointing downwards). 
    The same graph as before, but now with a second parabola shown in the third quadrant, pointing downwards. Dotted lines, horizontal and vertical respectively, connect the point P to the turning points of the first and second parabolas. An arrow indicating the 90 degree rotation is shown in the fourth quadrant.
  • The leading term of the quadratic will probably be $-x^2$, looking at my sketch and looking at the options.
  • Then, for the turning point to be at $(-2,-1)$, it will need to be $y=-(x+2)^2-1$.
  • Expand that out for $y=-x^2-4x-5$.
  • The answer is B.

 

Extension

  • Check your understanding; suppose that the original curve was $x=(y-2)^2+2$ instead.
  • You don't have to know this, but in general the way to rotate a graph by 90 degrees clockwise about the origin is to replace $x$ with $(-y)$ and, simultaneously, replace $y$ with $x$. How can you use this fact to do this question, where the rotation is not about the origin?

 

TMUA 2022 Paper 1 Question 10

  • A sequence of translations is just one big translation... or perhaps I should say one big translation parallel to the $x$-axis and one big translation parallel to the $y$-axis.
  • So after all of that, the graph $y=x^3$ will be transformed to $y=(x-a)^3+b$.
  • That's everything that I can make with translations!
  • I'd like to multiply that out and compare it to the graphs in the question. If they're not in that form, then I can't make them with my translations (if they are then I probably can, but I will be careful and make sure that I can solve for $a$ and $b$)
  • Multiplying out gives $y=x^3-3ax^2+3a^2x-a^3+b$.
  • So III is out straight away, because the coefficient of $x^3$ doesn't match.
  • For I, to get the coefficients to match, we would need $-3=-3a$ and $9=3a^2$ and also something involving $-27$ and $b$. But there's no value of $a$ that satisfies both of those equations, so no translation will give this graph.
  • For II, to get the coefficients to match, we would need $-9=-3a$ and $27=3a^2$... so we're off to a good start because $a=3$ works for both of those... and also $-3=-a^3+b$. I could solve for $b=24$ if I wanted to.
  • So I can get the graph of II (translate 3 units parallel to the $x$-axis and then $24$ units parallel to the $y$-axis).
  • The answer is C.

 

Extension

  • An alternative method that I considered when I read this question was "look to see if there is a point with both $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=0$ and $\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=0$". Would this work? If yes, retry the question with this method. If not, refine this method and then retry the question.
  • Find a necessary and sufficient condition involving $A$ and/or $B$ and/or $C$ such that there is a sequence of translations that takes the graph $y=x^3$ to the graph \[y=x^3+Ax^2+Bx+C.\]
  • Find a sequence of translations and/or stretches parallel to the axes that takes the graph $y=x^3$ to graph III.
  • Explain why there is no sequence of translations and/or stretches parallel to the axes that takes the graph $y=x^3$ to graph I.
  • Find necessary and sufficient conditions involving $A$ and/or $B$ and/or $C$ and/or $D$ such that there is a sequence of translations and/or stretches parallel to the axes that takes the graph $y=x^3$ to the graph \[y=Dx^3+Ax^2+Bx+C.\]

 

TMUA 2022 Paper 1 Question 18

  • I started by sketching $y=\mathrm{f}(x)$. This polynomial has degree 5, and it has repeated roots at $x=0$ and at $x=1$. There is a root at $x=2$. The values are positive for $x>2$ only. 
    A graph of a polynomial. The values are negative for almost, but not all, values of x. The values start very negative, increase to zero at the origin, where there's a repeated root, then the values are negative for a bit before another repeated root, then really quite negative again before finally increasing up through a final root, and then taking positive values and quickly rising off the top of the plot.
  • I don't want to work out where the turning points are precisely, so I'll just draw something with the roots in the right place, and I'll try to remember that I haven't checked the size of the values of $\mathrm{f}(x)$ away from the $x$-axis.
  • Next I sketched $y=\mathrm{g}(x)$. I can see that changing $p$ will just stretch the graph parallel to the $y$-axis, so I'll ignore that for now. This polynomial has degree 4 and has repeated roots at $x=q$ and at $x=r$. 
    A quartic with negative values. The graph starts very negative, increases to a turning point on the x-axis, then decreases for some more negative values, before increasing to a second turning point, also on the x-axis, before then plummeting back down to very negative values of x. The turning points are labelled q and r.
  • I'm interested in places where the graphs cross. Thinking about $x<0$, I'd like to know whether I'll get a crossing there. I know that $\mathrm{f}(0)=0$ and $\mathrm{g}(0)<0$, so the question is whether $\mathrm{f}(x)$ "catches up" and becomes more negative than $\mathrm{g}(x)$ as $x$ gets very negative. This will happen, because the degree of $\mathrm{f}$ is larger.
  • What is the "greatest number of distinct real solutions"? I suppose that the equation $\mathrm{f}(x)=\mathrm{g}(x)$ can be rearranged to a polynomial equation $\mathrm{h}(x)=0$ where $\mathrm{h}(x)=\mathrm{f}(x)-\mathrm{g}(x)$. That has degree 5, so the equation has at most 5 real solutions.
  • Can I choose $q$ and $r$ to guarantee four more distinct real solutions? Yes, if I make the graphs cross a lot!
  • If I put the turning point of $\mathrm{g}(x)$ at $q$ between $x=0$ and $x=1$, then I'll get two solutions, one between $0$ and $q$, and one between $q$ and $1$. Similarly, if I put the other turning point of $\mathrm{g}(x)$ between $x=1$ and $x=2$, then I'll get two more solutions over there.
  • This works for all $p$ because, for any value of $p$, the graphs really will cross in those intervals. In the cases given for the other options, changing $p$ might move the graph of $\mathrm{g}(x)$ away from the graph of $\mathrm{f}(x)$ in that region.
  • The answer is B.

 

Extension

  • For option C, find the minimum number of distinct real solutions as $p$ changes.

 

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