How to use this bank
- Start at Level 1 and move up only when the method feels automatic.
- Write full working on paper before opening any solution.
- Repeat the question types you still confuse.
- Use the quiz after this page.
Coverage map
Read-off, inverse checks, hyperbolas, quadratics, logarithms, exponentials, transformations, and mixed graph questions.
Level 1 | Knowledge
Foundation read-off questions
Identify the graph family, read the fixed features, and recall the inverse rule before the algebra gets heavier.
Read-off
Feature recall
Function behaviour
Correct notation
20%
Q1
Equation Station SA Practice Question
4 marks
Hyperbola
Asymptotes
Symmetry
Hyperbola read-off
Given \(f(x)=\dfrac{-2}{x+2}+2\), write down the equations of the asymptotes and the equation of the symmetry line with negative gradient.
Show solution
Memo route
- The graph is in the form \(\displaystyle y=\frac{a}{x-p}+q\), so \(p=-2\) and \(q=2\).
- The vertical asymptote is \(x=p\), so \(x=-2\).
- The horizontal asymptote is \(y=q\), so \(y=2\).
- The symmetry line with negative gradient passes through \((-2,2)\) with slope \(-1\): \(y-2=-(x+2)\).
Final answers
\[x=-2,\qquad y=2,\qquad y=-x\]
Q2
Equation Station SA Practice Question
6 marks
Quadratic
Turning point
Range
Quadratic read-off
Given \(f(x)=(x-1)^2-9\), determine the turning point, the equation of the axis of symmetry, the range, and the y-intercept.
Show solution
Memo route
- The function is already in turning-point form \(y=(x-p)^2+q\).
- So the turning point is \((1,-9)\).
- The axis of symmetry is \(x=1\).
- The graph opens upward, so the smallest y-value is \(-9\). Therefore the range is \(y\ge -9\).
- For the y-intercept, let \(x=0\): \(f(0)=(-1)^2-9=1-9=-8\).
Final answers
\[\text{Turning point }(1,-9),\quad x=1,\quad y\ge -9,\quad (0,-8)\]
Q3
Equation Station SA Practice Question
4 marks
Logarithm
Intercepts
Basic point
Logarithm read-off
Given \(k(x)=\log_{\frac12}x\), determine the x-intercept, the equation of the vertical asymptote, and one other point that lies on the graph.
Show solution
Memo route
- For the x-intercept, let \(y=0\): \(\log_{\frac12}x=0\).
- This gives \(x=1\), so the x-intercept is \((1,0)\).
- A logarithmic graph has vertical asymptote \(x=0\).
- Choose an easy point, for example \(x=2\): \(\log_{\frac12}2=-1\), so \((2,-1)\) lies on the graph.
Final answers
\[(1,0),\qquad x=0,\qquad (2,-1)\]
Q4
Equation Station SA Practice Question
5 marks
Straight line
Inverse
Domain and range
Inverse of a straight line
For the linear function \(f(x)=2x+3\), determine \(f^{-1}(x)\). Then state the domain and range of both \(f\) and \(f^{-1}\).
Show solution
Memo route
- Write \(y=2x+3\).
- Swap \(x\) and \(y\): \(x=2y+3\).
- Solve for \(y\): \(x-3=2y\), so \(y=\dfrac{x-3}{2}\).
- A non-horizontal line has domain \(\mathbb{R}\) and range \(\mathbb{R}\).
- The inverse swaps domain and range, so \(f^{-1}\) also has domain \(\mathbb{R}\) and range \(\mathbb{R}\).
Final answers
\[f^{-1}(x)=\frac{x-3}{2}\]
\[\operatorname{Dom}(f)=\operatorname{Ran}(f)=\operatorname{Dom}(f^{-1})=\operatorname{Ran}(f^{-1})=\mathbb{R}\]
Q5
Equation Station SA Practice Question
4 marks
Exponential
Read-off
Behaviour
Exponential read-off
Given \(h(x)=\left(\dfrac12\right)^x\), write down the y-intercept, the horizontal asymptote, whether the graph increases or decreases, and one other point on the graph.
Show solution
Memo route
- For the y-intercept, let \(x=0\): \(\left(\dfrac12\right)^0=1\), so the y-intercept is \((0,1)\).
- An exponential graph of this form has horizontal asymptote \(y=0\).
- Because the base \(\dfrac12\) lies between \(0\) and \(1\), the graph decreases from left to right.
- Choose an easy point, for example \(x=1\): \(\left(\dfrac12\right)^1=\dfrac12\), so \((1,\dfrac12)\) lies on the graph.
Final answers
\[(0,1),\qquad y=0,\qquad \text{decreasing},\qquad \left(1,\frac12\right)\]
Q6
Equation Station SA Practice Question
4 marks
Function test
Inverse
One-to-one
Identify functions with functional inverses
Without sketching graphs, consider \(f(x)=2x-1\), \(g(x)=x^2\), and \(h(x)=3^x\). State which of these are one-to-one. Hence state which of them have inverses that are functions on their natural domains.
Show solution
Memo route
- The line \(f(x)=2x-1\) is one-to-one because different x-values always give different y-values.
- The quadratic \(g(x)=x^2\) is not one-to-one on its full natural domain because, for example, \(g(2)=g(-2)\).
- The exponential \(h(x)=3^x\) is one-to-one because it is strictly increasing.
- Therefore the inverses of \(f\) and \(h\) are functions, but the inverse of \(g\) is not a function on its full natural domain.
Final answers
\[\text{One-to-one: }f\text{ and }h\]
\[\text{Inverse is a function on the natural domain: }f\text{ and }h\text{ only}\]
Level 2 | Routine procedures
Core exam-standard procedures
Calculate intercepts, build equations from graph facts, transform familiar graphs, and handle inverse restrictions cleanly.
Intercepts
Equation building
Restriction logic
Transformations
35%
Q7
Equation Station SA Practice Question
8 marks
Hyperbola
Intercepts
Symmetry line
Determine the features of a hyperbola
Given \(f(x)=\dfrac{2}{x-1}+2\), determine the asymptotes, the x-intercept, the y-intercept, and the equation of the symmetry line with negative gradient.
Show solution
Memo route
- From the form \(\displaystyle \frac{a}{x-p}+q\), the asymptotes are \(x=1\) and \(y=2\).
- For the x-intercept, let \(y=0\): \(0=\dfrac{2}{x-1}+2\).
- \(-2=\dfrac{2}{x-1}\Rightarrow -1=\dfrac{1}{x-1}\Rightarrow x-1=-1\Rightarrow x=0\).
- So the x-intercept is \((0,0)\).
- For the y-intercept, let \(x=0\): \(f(0)=\dfrac{2}{-1}+2=0\), so the y-intercept is also \((0,0)\).
- The centre is \((1,2)\). The symmetry line with negative gradient is \(y-2=-(x-1)\).
Final answers
\[x=1,\qquad y=2,\qquad (0,0),\qquad y=-x+3\]
Q8
Equation Station SA Practice Question
8 marks
Quadratic
Turning-point form
Sign of graph
Determine a quadratic from given information
A parabola has turning point \((2,10)\) and y-intercept \((0,8)\). Determine its equation. Then, for \(g(x)=x^2-6x+8\), determine the turning point and the values of \(x\) for which \(g(x)<0\).
Show solution
Memo route
- Use turning-point form: \(f(x)=a(x-2)^2+10\).
- Substitute the y-intercept: \(8=a(0-2)^2+10\).
- \(8=4a+10\Rightarrow 4a=-2\Rightarrow a=-\dfrac12\).
- So \(\displaystyle f(x)=-\frac12(x-2)^2+10\).
- For \(g(x)=x^2-6x+8\), the turning point has x-value \(\dfrac{-(-6)}{2}=3\).
- \(g(3)=9-18+8=-1\), so the turning point is \((3,-1)\).
- Factor \(g(x)\): \(x^2-6x+8=(x-2)(x-4)\).
- Therefore \(g(x)<0\) between the roots, so \(2<x<4\).
Final answers
\[f(x)=-\frac12(x-2)^2+10\]
\[\text{Turning point of }g=(3,-1),\qquad g(x)<0\text{ for }2<x<4\]
Q9
Equation Station SA Practice Question
6 marks
Inverse
Quadratic
Restriction
Inverse of a quadratic relation
Given \(f(x)=2x^2\), determine the inverse relation in the form \(y=\ldots\). Explain why the inverse is not a function on the full domain. Then write the inverse if the original domain is restricted to \(x\le 0\).
Show solution
Memo route
- Write \(y=2x^2\) and swap \(x\) and \(y\): \(x=2y^2\).
- Solve for \(y\): \(\displaystyle y=\pm\sqrt{\frac{x}{2}}\).
- This inverse relation is not a function because one x-value gives two y-values, one positive and one negative.
- If the original graph is restricted to \(x\le 0\), the inverse must return the negative branch.
Final answers
\[y=\pm\sqrt{\frac{x}{2}}\]
\[\text{Restricted inverse: }f^{-1}(x)=-\sqrt{\frac{x}{2}}\]
Q10
Equation Station SA Practice Question
5 marks
Logarithm
Inverse
Transformation
Inverse of a logarithmic function
Let \(k(x)=\log_{\frac12}x\). Determine \(k^{-1}(x)\). If \(h(x)=k^{-1}(x+2)\), determine the y-intercept of \(h\) and the horizontal asymptote of \(h\).
Show solution
Memo route
- The inverse of \(\log_{\frac12}x\) is the exponential function \(\displaystyle k^{-1}(x)=\left(\frac12\right)^x\).
- So \(\displaystyle h(x)=k^{-1}(x+2)=\left(\frac12\right)^{x+2}\).
- For the y-intercept, let \(x=0\): \(\displaystyle h(0)=\left(\frac12\right)^2=\frac14\).
- This is still an exponential graph, so the horizontal asymptote is \(y=0\).
Final answers
\[k^{-1}(x)=\left(\frac12\right)^x,\qquad h(0)=\frac14,\qquad y=0\]
Q11
Equation Station SA Practice Question
5 marks
Exponential
Transformation
Asymptote
Determine the transformed exponential
Start with \(g(x)=2^x\). A new graph \(t\) is obtained by reflecting \(g\) in the y-axis and then shifting the result 1 unit downward. Write down the equation of \(t\), the y-intercept of \(t\), and the horizontal asymptote of \(t\).
Show solution
Memo route
- Reflect in the y-axis by replacing \(x\) with \(-x\): \(g(-x)=2^{-x}=\left(\dfrac12\right)^x\).
- Shift the reflected graph 1 unit downward: \(\displaystyle t(x)=2^{-x}-1=\left(\frac12\right)^x-1\).
- For the y-intercept, let \(x=0\): \(t(0)=2^0-1=0\), so the y-intercept is \((0,0)\).
- The basic exponential has horizontal asymptote \(y=0\). After shifting down 1 unit, the new asymptote is \(y=-1\).
Final answers
\[t(x)=2^{-x}-1=\left(\frac12\right)^x-1,\qquad (0,0),\qquad y=-1\]
Q12
Equation Station SA Practice Question
5 marks
Quadratic
Transformation
Inverse
Quadratic transformation and inverse
Let \(f(x)=(x-1)^2-9\) and \(g(x)=x^2\). Describe the transformation that maps \(f\) to \(g\). Then write down the inverse of \(g\) if the domain of \(g\) is restricted to \(x\ge 0\).
Show solution
Memo route
- The graph \(f(x)=(x-1)^2-9\) is the basic parabola shifted 1 unit right and 9 units down.
- To move from \(f\) back to \(g(x)=x^2\), shift the graph 1 unit left and 9 units up.
- For the inverse of \(g(x)=x^2\) with \(x\ge 0\), write \(y=x^2\), swap \(x\) and \(y\), and solve.
- This gives \(x=y^2\Rightarrow y=\pm\sqrt{x}\). Because the original domain is \(x\ge 0\), choose the positive branch.
Final answers
\[\text{Shift }1\text{ unit left and }9\text{ units up}\]
\[g^{-1}(x)=\sqrt{x}\]
Level 3 | Complex procedures
Linked graph reasoning
Connect graph features, inverse rules, gradients, comparisons, and reflections inside one question.
Compare graphs
Average gradient
Reflection logic
Linked transformations
30%
Q13
Equation Station SA Practice Question
12 marks
Quadratic
Line comparison
Inequality
Comparison of a line and a parabola
The line \(g(x)=-3x+20\) intersects a parabola \(f\) at \(A\) and \(B(6,2)\). Point \(A\) is the turning point of \(f\) and has x-coordinate \(3\). Determine the y-coordinate \(k\) of \(A\), the range of \(y=-f(x)\), the values of \(a\), \(b\) and \(c\) if \(f(x)=ax^2+bx+c\), and the values of \(x\) for which \(f(x)>g(x)\).
Show solution
Memo route
- Because \(A\) lies on \(g\) and its x-value is \(3\), \(k=g(3)=-3(3)+20=11\). So \(A=(3,11)\).
- Use turning-point form for \(f\): \(f(x)=a(x-3)^2+11\).
- Substitute \(B(6,2)\): \(2=a(6-3)^2+11=9a+11\).
- \(9a=-9\Rightarrow a=-1\), so \(f(x)=-(x-3)^2+11\).
- Expand: \(f(x)=-x^2+6x+2\). Therefore \(a=-1\), \(b=6\), \(c=2\).
- The graph of \(-f(x)\) opens upward and has minimum y-value \(-11\), so its range is \(y\ge -11\).
- For \(f(x)>g(x)\): \(-x^2+6x+2>-3x+20\).
- \(-x^2+9x-18>0\Rightarrow x^2-9x+18<0\Rightarrow (x-3)(x-6)<0\).
- So \(f(x)>g(x)\) for \(3<x<6\).
Final answers
\[k=11,\qquad y\ge -11,\qquad f(x)=-x^2+6x+2,\qquad 3<x<6\]
Q14
Equation Station SA Practice Question
7 marks
Exponential
Inverse
Inequality
Exponential inverse inequality
Given \(f(x)=b^x\) and \(A(3,8)\) lies on the graph of \(f\), determine \(b\), determine \(f^{-1}(x)\), and solve the inequality \(f^{-1}(x)<4\).
Show solution
Memo route
- Substitute the point: \(8=b^3\), so \(b=2\).
- Therefore \(f(x)=2^x\).
- The inverse is \(\displaystyle f^{-1}(x)=\log_2 x\).
- Solve \(\log_2 x<4\). Since the base is greater than \(1\), the inequality direction stays the same.
- So \(x<2^4=16\). Also the logarithm only exists for \(x>0\).
Final answers
\[b=2,\qquad f^{-1}(x)=\log_2 x,\qquad 0<x<16\]
Q15
Equation Station SA Practice Question
8 marks
Hyperbola
Reflection
Symmetry line
Reflection in a hyperbola symmetry line
Consider \(f(x)=\dfrac{3}{x-2}+1\). Determine the asymptotes, determine the symmetry line with positive gradient, and then find the image of \(R(1,-2)\) when it is reflected in that symmetry line.
Show solution
Memo route
- From the form \(\displaystyle \frac{a}{x-p}+q\), the asymptotes are \(x=2\) and \(y=1\).
- The centre is \((2,1)\).
- The symmetry line with positive gradient has slope \(1\) and passes through \((2,1)\): \(y-1=x-2\).
- So the required line is \(y=x-1\).
- To reflect across \(y=x-1\), shift up by \(1\), reflect across \(y=x\), then shift back down by \(1\).
- \((1,-2)\) becomes \((1,-1)\), then \((-1,1)\), then \((-1,0)\).
Final answers
\[x=2,\qquad y=1,\qquad y=x-1,\qquad R'(-1,0)\]
Q16
Equation Station SA Practice Question
5 marks
Logarithm
Inverse
Shift
Logarithm to inverse transformation
Point \(B(4,2)\) lies on \(g(x)=\log_m x\). Determine the value of \(m\). Then determine \(g^{-1}(x)\). If \(h\) is obtained by shifting \(g^{-1}(x)\) 2 units to the left, write down the equation of \(h\).
Show solution
Memo route
- Because \(B(4,2)\) lies on \(g\), \(\log_m 4=2\).
- This means \(m^2=4\). Since a logarithm base must be positive and not \(1\), \(m=2\).
- So \(g(x)=\log_2 x\).
- The inverse is \(g^{-1}(x)=2^x\).
- A shift 2 units left replaces \(x\) with \(x+2\), so \(h(x)=2^{x+2}\).
Final answers
\[m=2,\qquad g^{-1}(x)=2^x,\qquad h(x)=2^{x+2}\]
Q17
Equation Station SA Practice Question
4 marks
Average gradient
Parabola
Comparison
Average gradient on a parabola
The line \(g(x)=-3x+20\) and the parabola \(f(x)=-(x-3)^2+11\) intersect at \(A(3,11)\) and \(B(6,2)\). Determine the average gradient of \(f\) between \(A\) and \(B\). Then compare it with the gradient of \(g\).
Show solution
Memo route
- Use the average-gradient formula between the two points on \(f\): \(\displaystyle \frac{y_2-y_1}{x_2-x_1}\).
- Substitute \(A(3,11)\) and \(B(6,2)\): \(\displaystyle \frac{2-11}{6-3}=\frac{-9}{3}=-3\).
- The gradient of the line \(g(x)=-3x+20\) is also \(-3\).
- So the average gradient of \(f\) between \(A\) and \(B\) is equal to the gradient of \(g\).
Final answers
\[\text{Average gradient of }f=-3,\qquad \text{and it is equal to the gradient of }g\]
Q18
Equation Station SA Practice Question
6 marks
Exponential
Transformation
Inequality
Transformation followed by inequality
Let \(h(x)=2^x\) and \(t(x)=\left(\dfrac12\right)^x-1\). Describe the transformation that maps \(h\) to \(t\). Then solve \(t(x)>-\dfrac34\).
Show solution
Memo route
- Replacing \(x\) with \(-x\) reflects \(h(x)=2^x\) in the y-axis and gives \(2^{-x}=\left(\dfrac12\right)^x\).
- Subtracting \(1\) shifts the reflected graph 1 unit downward, so that is the full transformation.
- Solve the inequality: \(\left(\dfrac12\right)^x-1>-\dfrac34\Rightarrow \left(\dfrac12\right)^x>\dfrac14\).
- Since \(\dfrac14=\left(\dfrac12\right)^2\) and the base is between \(0\) and \(1\), the function decreases, so \(x<2\).
Final answers
\[\text{Reflect in the y-axis, then shift 1 unit downward}\]
\[x<2\]
Level 4 | Problem solving
Mixed mastery questions
Choose the method yourself and connect roots, inverse branches, sign logic, and transformations with confidence.
Mixed sign logic
Branch choice
Reflections
One-to-one decisions
15%
Q19
Equation Station SA Practice Question
8 marks
Quadratic
Logarithm
Sign analysis
Sign analysis with a parabola and logarithm
A parabola \(f\) has axis of symmetry \(x=2\), x-intercept \(A(5,0)\), and point \(B(4,2)\) lies on both \(f\) and \(g(x)=\log_m x\). Determine \(m\), determine the equation of \(f\), and determine the values of \(x\) for which \(f(x)\,g(x)>0\).
Show solution
Memo route
- Because \(B(4,2)\) lies on \(g(x)=\log_m x\), \(\log_m 4=2\), so \(m=2\).
- The axis of symmetry of \(f\) is \(x=2\). Since one root is \(x=5\), the other root is \(x=-1\).
- Therefore \(f(x)=a(x-5)(x+1)\).
- Use \(B(4,2)\): \(2=a(4-5)(4+1)=a(-1)(5)\Rightarrow a=-\dfrac25\).
- So \(\displaystyle f(x)=-\frac25(x-5)(x+1)\).
- On the domain of \(g\), we need \(x>0\). Also \(g(x)=\log_2 x\) is negative on \(0<x<1\), zero at \(x=1\), and positive for \(x>1\).
- The parabola \(f\) is positive between its roots, so on \(x>0\) it is positive for \(0<x<5\) and negative for \(x>5\).
- The product \(f(x)g(x)\) is positive only when both are positive, which happens on \(1<x<5\).
Final answers
\[m=2,\qquad f(x)=-\frac25(x-5)(x+1),\qquad 1<x<5\]
Q20
Equation Station SA Practice Question
6 marks
Inverse
Quadratic
Branch choice
Choose the correct inverse branch
Given \(f(x)=-6x^2\), write a restriction on the domain of \(f\) so that the inverse is a function and \(f^{-1}(x)\le 0\). Then determine \(f^{-1}(x)\) and state the domain of the inverse.
Show solution
Memo route
- To make the inverse output non-positive values, the original graph must be restricted to the left-hand branch, so \(x\le 0\).
- Write \(y=-6x^2\) and swap \(x\) and \(y\): \(x=-6y^2\).
- Solve for \(y\): \(\displaystyle y=\pm\sqrt{\frac{-x}{6}}\).
- Because \(f^{-1}(x)\le 0\), choose the negative branch.
- The inverse exists only for \(x\le 0\), because the outputs of \(-6x^2\) are never positive.
Final answers
\[\text{Restrict }f\text{ to }x\le 0,\qquad f^{-1}(x)=-\sqrt{\frac{-x}{6}},\qquad \operatorname{Dom}(f^{-1})=\{x:x\le 0\}\]
Q21
Equation Station SA Practice Question
6 marks
Exponential
Reflection
Inverse
Reflect and then invert an exponential
A point \((2,9)\) lies on \(f(x)=a^x\). Determine \(a\). If \(g\) is obtained by reflecting \(f\) in the y-axis, determine \(g(x)\), determine \(g^{-1}(x)\), and state the domain of \(g^{-1}\).
Show solution
Memo route
- Substitute the point into \(f(x)=a^x\): \(9=a^2\).
- For an exponential base, \(a>0\), so \(a=3\).
- Reflecting in the y-axis means replace \(x\) with \(-x\): \(g(x)=f(-x)=3^{-x}=\left(\dfrac13\right)^x\).
- The inverse of \(\left(\dfrac13\right)^x\) is \(\log_{\frac13}x\).
- A logarithm only exists for positive inputs, so the domain of \(g^{-1}\) is \(x>0\).
Final answers
\[a=3,\qquad g(x)=3^{-x}=\left(\frac13\right)^x,\qquad g^{-1}(x)=\log_{\frac13}x,\qquad x>0\]
Q22
Equation Station SA Practice Question
5 marks
Logarithm
Inverse
Inequality
Logarithmic inequality with base smaller than 1
Let \(f(x)=\left(\dfrac13\right)^x\), so \(f^{-1}(x)=\log_{\frac13}x\). Determine the values of \(x\) for which \(f^{-1}(x)\ge -3\).
Show solution
Memo route
- Start with \(\log_{\frac13}x\ge -3\).
- The logarithm is only defined for \(x>0\).
- Because the base \(\dfrac13\) is between \(0\) and \(1\), the function is decreasing, so the inequality reverses when we rewrite it exponentially.
- Therefore \(x\le \left(\dfrac13\right)^{-3}=27\).
- Combine this with the domain restriction \(x>0\).
Final answers
\[0<x\le 27\]
Q23
Equation Station SA Practice Question
8 marks
Inverse
One-to-one
Restriction
Decide whether a restriction is needed
For each function, decide whether its inverse is a function on the stated domain. If it is, write the inverse. If it is not, give one suitable restriction on the original domain that makes the inverse a function, and then write that inverse: \(p(x)=3x-4\), \(q(x)=x^2-1\) on \(\mathbb{R}\), and \(r(x)=2^x\).
Show solution
Memo route
- \(p(x)=3x-4\) is a straight line with non-zero gradient, so it is one-to-one and its inverse is a function.
- Write \(y=3x-4\), swap \(x\) and \(y\), and solve: \(x=3y-4\Rightarrow y=\dfrac{x+4}{3}\).
- \(q(x)=x^2-1\) on \(\mathbb{R}\) is not one-to-one, so its inverse on the full domain is not a function.
- One suitable restriction is \(x\ge 0\). Then \(y=x^2-1\Rightarrow x=y^2-1\Rightarrow y=\sqrt{x+1}\).
- \(r(x)=2^x\) is one-to-one on its natural domain, so its inverse is the logarithmic function \(r^{-1}(x)=\log_2 x\).
Final answers
\[p^{-1}(x)=\frac{x+4}{3}\]
\[q(x)=x^2-1\text{ needs a restriction such as }x\ge 0,\qquad q^{-1}(x)=\sqrt{x+1}\]
\[r^{-1}(x)=\log_2 x\]
Q24
Equation Station SA Practice Question
8 marks
Hyperbola
Parameters
Symmetry
Determine a hyperbola from its structure
A hyperbola has equation \(f(x)=\dfrac{a}{x-p}+q\). Its domain is \(x>1\), one line of symmetry is \(y=x-3\), and the graph passes through \((2,2)\). Determine \(p\), \(q\), and \(a\), and hence write down the equation of \(f\).
Show solution
Memo route
- The domain excludes \(x=p\). Since the domain is \(x>1\), the vertical asymptote is \(x=1\), so \(p=1\).
- The positive-gradient symmetry line of a hyperbola in this form passes through the centre \((p,q)\) and has equation \(y-q=x-p\).
- So its equation can be written \(y=x+(q-p)\). Given \(y=x-3\), we get \(q-p=-3\).
- Substitute \(p=1\): \(q-1=-3\Rightarrow q=-2\).
- Use the point \((2,2)\): \(2=\dfrac{a}{2-1}-2\Rightarrow 2=a-2\Rightarrow a=4\).
Final answers
\[p=1,\qquad q=-2,\qquad a=4\]
\[f(x)=\frac{4}{x-1}-2\]
Q25
Equation Station SA Practice Question
7 marks
Logarithm
Graph interpretation
No equation given
Determine the equation from a sketch (no equation given)
The graph of \(f(x)=\log_b(x-p)+q\) is sketched below with its vertical asymptote and two points labelled (no equation given). Determine the values of \(p\), \(b\) and \(q\), and hence write down the equation of \(f\).
Show solution
Memo route
- The vertical asymptote is \(x=p\). The sketch shows \(x=2\), so \(p=2\).
- Substitute the x-intercept \((3,0)\): \(0=\log_b(3-2)+q\Rightarrow0=\log_b(1)+q\). Since \(\log_b(1)=0\) for any base, \(q=0\).
- Substitute \((11,2)\) with \(p=2,q=0\): \(2=\log_b(11-2)\Rightarrow2=\log_b(9)\Rightarrow b^2=9\Rightarrow b=3\) (a logarithmic base must be positive).
Final answers
\[p=2,\qquad b=3,\qquad q=0\]
\[f(x)=\log_3(x-2)\]
Next step after the bank
When these feel solid, move to the quiz.