Graphs of Functions Worked ExamplesIGCSE Maths 0580

(a) Set $y = 0$ and factorise, taking out the common factor first.

Set each factor to zero.

(b) The $x^3$ coefficient is positive, so the curve falls from bottom-left and rises to top-right.

Check a value to confirm: at $x = 3$, $y = 27 - 12 = 15$, which is positive as expected.

(a) The function is undefined at $x = 0$, giving a vertical asymptote there; the value never reaches zero, giving a horizontal one.

(b) Because the curve cannot cross $x = 0$, it is split into the part where $x > 0$ and the part where $x < 0$.

(c) Replacing $x$ by $-x$ gives $-y$, so the curve maps onto itself under a half-turn about the origin.

At an intersection both equations give the same $y$, so set the right-hand sides equal.

Rearrange so the quadratic equals zero.

Factorise.

Substitute into the linear equation — it is simpler and less error-prone.

Check in the quadratic: $1 - 4 + 3 = 0$ ✓ and $16 - 16 + 3 = 3$ ✓.

(a) Set $x = 0$. Any non-zero base to the power $0$ is $1$.

(b) Evaluate at each value.

(c) As $x$ becomes large and negative the value approaches zero but never reaches it.

(a) The equation asks where $y = 0$, which is where the curve meets the $x$-axis. Read off those $x$-values.

(b) This asks where $y = 2$, so draw the horizontal line $y = 2$ across the graph.

Read the $x$-coordinates where that line cuts the curve — there are two, since the line lies above the minimum of $-1$.

The drawn curve is fixed, so rearrange the target equation until one side is exactly $x^2 - 4x + 3$.

Add $x$ to both sides to turn $-5x$ into $-4x$.

Add $2$ to both sides to turn the constant $1$ into $3$.

The left side now matches the drawn curve, so the required line is the right side.

A is an upside-down U crossing the $y$-axis at $2$. Of the two quadratics, only $y = 2 - x^2$ has a negative $x^2$ term (so it opens downwards) and a $y$-intercept of $2$.

B is a reciprocal graph with branches in the top-left and bottom-right quadrants, so $y$ is negative when $x$ is positive. Check $x = 1$: $-\frac{3}{1} = -3$, below the axis. ✓

C crosses the $y$-axis at $1$ and falls from left to right, getting closer to the $x$-axis. $y = 2^x$ rises, so C is the other exponential. Check $x = 1$: $2^{-1} = \frac{1}{2}$, less than $1$. ✓

D is a cubic through $-1$, $0$ and $1$ that falls away on the right, so its $x^3$ term is negative. Check $x = 2$: $2 - 8 = -6$, below the axis. ✓

(a) Read the lowest point of the curve.

(b) Draw the horizontal line $y = 6$. It meets the curve twice. The second point is exactly on grid lines: $2^2 + \frac{4}{2} = 6$ ✓.

(c)(i) Where the line meets the curve, the two expressions for $y$ are equal.

Multiply every term by $x$ (which is not zero), then collect everything on one side.

(ii) Plot the line using points such as $(0.5, 4)$, $(2, 7)$ and $(4, 11)$. It crosses the curve twice. Check $x = 1$ in the cubic: $1 - 2 - 3 + 4 = 0$ ✓.

(iii) The third solution is negative, but the graph has only been drawn for $0.5 \leqslant x \leqslant 4$, so that intersection is not on the grid.

(a) Both turning points sit on grid crossings. Check: $f(1) = 1 - 6 + 9 = 4$ and $f(3) = 27 - 54 + 27 = 0$.

(b) The solutions of $f(x) = k$ are where the horizontal line $y = k$ meets the curve. (i) $y = 2$ lies between the two turning-point heights, so it crosses all three sections of the curve.

(ii) $y = 4$ just touches the curve at the maximum $(1, 4)$ and crosses it again at $x = 4$.

(iii) $y = -1$ is below the minimum, so it meets only the left-hand part of the curve.

(c) There are three crossings only when the line lies strictly between the turning-point heights. At $k = 0$ or $k = 4$ it touches a turning point and there are only two.

(d) Rearrange so one side is exactly $f(x)$: adding $x$ to both sides turns $8x$ into $9x$. The line to draw is $y = x$.

The line $y = x$ meets the curve at $(0, 0)$, $(2, 2)$ and $(4, 4)$.

Confirm by factorising.

(a) Read where the left-hand branch meets the $x$-axis. Check it: at $y = 0$, $\frac{3}{x} = -2$, so $x = -1.5$.

(b) $\frac{3}{x}$ is undefined at $x = 0$, so the curve never meets the $y$-axis. As $x$ gets large, $\frac{3}{x}$ gets close to $0$, so $y$ gets close to $2$. Both asymptotes are lines, so write them as equations.

(c) Rearrange the equation so that one side is exactly the curve already drawn, $\frac{3}{x} + 2$. Add $2x$ to both sides, then add $1$.

So draw the line $y = 2x + 1$ with a ruler, for example through $(-2, -3)$, $(0, 1)$ and $(3, 7)$.

Read the $x$-coordinates where the line crosses the curve: at $(-1, -1)$ on the left branch and $(1.5, 4)$ on the right branch.

Check in the original equation: $\frac{3}{-1} + 2 + 1 = 0$ ✓ and $\frac{3}{1.5} - 3 + 1 = 0$ ✓.

(a) Square the bracket first.

Multiply each term by $x$, then by $-4$, and collect like terms.

(b) Shape: the $x^3$ coefficient is positive, so the curve starts bottom-left and ends top-right.

Roots: $y = 0$ when $x = 4$ or $x = -1$. The factor $(x + 1)$ is squared, so $x = -1$ is a repeated root. The curve touches the $x$-axis there and turns back, rather than crossing it.

$y$-intercept: put $x = 0$, or read the constant term from part (a).

Put it together. From bottom-left the curve rises to touch the $x$-axis at $-1$ (a maximum), falls through $-4$ on the $y$-axis to a minimum between $0$ and $4$, then rises through $4$ on the $x$-axis. Label $-1$, $4$ and $-4$ on the sketch.

(a) Substitute each missing value, using a bracket for the negative. $(-1)^2 = +1$, so $-4(-1)^2 = -4$.

The other two values.

(b) Plot all seven points accurately and join them with one smooth curve — no ruled segments. The curve turns smoothly at the top near $(0, 5)$ and at the bottom between $x = 2$ and $x = 3$, going slightly below $-4$.

(c) Rearrange the equation so one side is exactly $x^3 - 4x^2 + 5$. Subtract $x$ and add $3$ to both sides.

Draw the ruled line $y = 3 - x$, for example through $(-2, 5)$, $(0, 3)$ and $(4, -1)$.

Read the $x$-coordinates of the three crossing points.

Check $x = 1$ in the equation: $1 - 4 + 1 + 2 = 0$ ✓. (The exact values of the other two roots are $\frac{3 \pm \sqrt{17}}{2}$, which are $-0.56$ and $3.56$.)