Graphs of Functions: past paper questions & worked solutions

This is a quadratic graph with a negative coefficient of \(x^2\), so it is a downward-opening parabola.

That is graph **B**.

This is an increasing cubic graph passing through the origin.

That is graph **C**.

(i) \[ y=-\frac1x \] has branches in quadrants II and IV, so it is **Graph E**.

(ii) \[ y=\sin x^\circ \] is the sine-shaped graph through the origin, so it is **Graph A**.

Point \(A\) is the maximum point.

So

Point \(B\) is the positive \(x\)-intercept after the minimum.

For this point,

So

Point \(A\) is the first positive \(x\)-intercept, so \(y=0\).

So

Point \(B\) is the minimum point.

The minimum value of cosine is \(-1\), which occurs when

So

The transformation *y* = *f*(4*x*) represents a horizontal stretch by a scale factor of $\\frac{1}{4}$. The x-coordinate is multiplied by $\\frac{1}{4}$ while the y-coordinate stays the same. $$\\text{New minimum} = \\left(8 \\times \\frac{1}{4}, 1\\right) = (2, 1)$$

(2,1)

**Step 1: Find the vertical shift *b*.** The standard *y* = cos (*x*) graph fluctuates between  − 1 and 1, centered at *y* = 0. The given graph fluctuates between a minimum of *y* = 1 and a maximum of *y* = 3. The center line is $\\frac{3 + 1}{2} = 2$. Therefore, *b* = 2.

**Step 2: Find the frequency multiplier *a*.** The period of the given graph completes one full cycle between *x* = 0 and *x* = 90^∘ (it starts at 3, dips to 1, and returns to 3 at 90^∘). The standard cosine period is 360^∘. $$\\text{Period} = \\frac{360^\\circ}{a}$$ $$90^\\circ = \\frac{360^\\circ}{a} \\implies a = 4$$

The completed table is:

Plot the points and join them with a smooth cubic curve.

From the graph,

The maximum is about \(45^\circ\). For a sine graph, a maximum happens when the angle inside the sine is \(90^\circ\), so

Use \(y=x^3+2x+3\).

For \(x=-3\),

For \(x=2\),

So the table is

Plot these points and draw a smooth cubic curve through them.

Estimate the gradient of *y* = *f*(*x*) at the point where *x* = 3.

Draw a tangent to the curve at *x* = 3. From the graph, a suitable tangent passes through approximately (1, −2) and (5, 6):

Calculate gradient of the tangent: $$\\text{Gradient} = \\frac{\\Delta y}{\\Delta x} = \\frac{6-(-2)}{5-1} = \\frac{8}{4} = 2$$

Answers in the range 1 to 3 are typically accepted. Draw the tangent triangle as large as possible for the most accurate estimate.

The graph of *y* = *a*sin (*x*+*b*)^∘ + *c* for 0 ≤ *x* ≤ 360 is shown. Find suitable values for *a*, *b*, and *c*.

Read the maximum and minimum values from the graph to find the **amplitude** *a* and **vertical shift** *c*: $$a = \\frac{\\text{max}-\\text{min}}{2} = \\frac{4-(-2)}{2} = 3 \\qquad c = \\frac{\\text{max}+\\text{min}}{2} = \\frac{4+(-2)}{2} = 1$$

Find the **phase shift** *b*. The standard sin  has its peak at *x* = 90°. From the graph, the maximum occurs at *x* ≈ 45°: *x* + *b* = 90° ⟹ 45° + *b* = 90° ⟹ *b* = 45°

Verify at *x* = 0: $y = 4\\sin(0+45)° = 4\\times\\frac{\\sqrt{2}}{2} \\approx 2.83$ — consistent with graph

The period of the graph is still 360° (unchanged), so no horizontal stretch is applied.

For \[ y=\frac{10}{x^2} \] the graph is positive on both sides and symmetrical in the \(y\)-axis.

This is Graph D.

For \[ y=x-3+3x^2-x^3 \] the leading term is \(-x^3\), so it is a cubic falling to the right with two turning points.

This is Graph C.

For \[ y=-\frac3x \] the graph is a negative reciprocal graph in quadrants 2 and 4.

This is Graph B.

\[ y=\sin x \] is the periodic sine-shaped graph, so it is **Graph C**.

\[ y=\frac{3}{x} \] is the positive reciprocal graph, so it is **Graph B**.

\[ y=4x^3-5x \] is a cubic graph with two turning points, so it is **Graph A**.

From the graph, the maximum value is

and the minimum value is

The amplitude is half the difference:

The midline is the average:

The distance between consecutive maxima is \(120^\circ\), so the period is

For \(y=a\cos bx^\circ+c\), the period is

So

\[ y=-2x+3 \] is a negative straight line with positive \(y\)-intercept, so it is **Graph C**.

\[ y=\frac1x \] is the reciprocal graph in quadrants I and III, so it is **Graph B**.

\[ y=\tan x \] has repeating branches with vertical asymptotes and passes through the origin, so it is **Graph D**.

\[ y=(x+1)(x-1)(x-2) \] is a positive cubic with three roots, so it is **Graph H**.

Draw a tangent to the curve at the point where \[ x=2 \]

Using two clear points on the tangent, for example approximately \((1.5,0.6)\) and \((2.3,2.2)\), \[ \text{gradient}\approx\frac{2.2-0.6}{2.3-1.5} \]

For \[ y=-\frac{2}{x^2} \] the graph is negative on both sides of the \(y\)-axis and is symmetrical in the \(y\)-axis.

This is Graph D.

For \[ y=-\frac12x^3 \] the graph is a decreasing cubic through the origin.

This is Graph B.

For \[ y=\frac5x \] the graph is a positive reciprocal graph in quadrants 1 and 3.

This is Graph A.

Use \(y=x+1/x\): \[y(0.5)=0.5+2=2.5,\quad y(1)=1+1=2,\quad y(4)=4+0.25=4.25,\quad y(5)=5+0.2=5.2.\]

Substitute each required x-value into the formula.

Plot \[(0.25,4.25),\ (0.5,2.5),\ (1,2),\ (2,2.5),\ (3,10/3),\ (4,4.25),\ (5,5.2).\]

Use x horizontally and y vertically on the supplied grid; 10/3 is approximately 3.33.

Join the points with one smooth curve for \(0.25\le x\le5\), descending to its minimum \((1,2)\) and then ascending.

The graph is curved, not a set of straight segments, and includes both endpoints.

The completed table is:

Plot the points and join them with a smooth curve.

The completed table is:

Plot the points and join them with a smooth U-shaped curve.

The completed table is:

Plot the points and join them with a smooth U-shaped curve.

The completed table is:

Plot the points and join them with a smooth decreasing reciprocal curve.

The completed table is:

Plot the points and join them with a smooth curve.

The completed table is:

Plot these points and join them with a smooth U-shaped curve.

The turning point is read directly from the completed square form:

For the \(y\)-intercept, put \(x=0\):

So the \(y\)-intercept is

For the \(x\)-intercepts, put \(y=0\):

So the \(x\)-intercepts are

Since \(a>4\), the turning point is below the \(x\)-axis and the curve crosses the \(x\)-axis twice.

At \(x=1\), the point is approximately \((1,4)\). Draw a tangent there, extending at least \(3\text{ cm}\). A suitable approximate tangent is \(y=-5x+9\), through \((0.5,6.5)\) and \((2,-1)\).

The gradient at a point is estimated by the gradient of its tangent; use two well-separated points on the line.

Calculate change in y divided by change in x using the axis values.

Rearrange so the given curve y=f(x) can be intersected by a straight line.

Draw \(y=2x-6\) through \((0,-6),(3,0),(4.5,3)\) on the same grid. Read the x-coordinates of its two intersections with the curve: approximately \(2.0\) and \(4.4\).

Each intersection has equal y-values on the line and curve and therefore solves the original equation.

The completed table is:

For \[ x+\frac4x=6 \] we need \[ \frac12\left(x+\frac4x\right)=3 \]

So draw the horizontal line \(y=3\) on the graph.

At the intersection points,

The two roots are \(p\) and \(q\).

For this quadratic,

Solve the two equations:

Use one root in \(y=x^2-8x-9\):

The completed table is:

The graph of \[ y=-\frac1{x^2} \] is below the \(x\)-axis on both sides and is symmetrical in the \(y\)-axis.

For part (a), solve

Using the graph gives approximately

For part (b), we need to solve

Multiply by 2:

Now

So draw the straight line

The intersections of this line with the curve give approximately

\(y=x^2-3x=x(x-3)\): U-shape through \(O\) and \(x=3\) → graph D.

\(y=-x^3+9x\): negative cubic through the origin → graph E.

Plot all 7 points and draw a smooth curve (no straight segments).

Draw a tangent to the curve at \(P\).

A reasonable tangent gives an estimated gradient of about \[ m\approx -0.6 \]

The point \(P\) is approximately \[ (2,2.4) \]

Using \(y=mx+c\), \[ 2.4\approx -0.6(2)+c \]

So an approximate tangent equation is \[ y\approx -0.6x+3.6 \]

The equation \(f(x)=k\) has exactly two different solutions when the horizontal line touches one turning level.

From the graph, these levels are approximately \[ k=3 \quad \text{and} \quad k=-1 \]