AQA C1 2009 January — Question 4 10 marks

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2009
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscriminant and conditions for roots
TypeProve/show always positive
DifficultyEasy -1.2 This is a routine C1 completing-the-square question with standard follow-up parts. Part (a) requires basic algebraic manipulation, part (b) involves reading off the minimum point and sketching, and part (c) asks for a standard transformation description. All techniques are textbook exercises with no problem-solving or novel insight required, making it easier than average.
Spec1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x)

4
    1. Express \(x ^ { 2 } + 2 x + 5\) in the form \(( x + p ) ^ { 2 } + q\), where \(p\) and \(q\) are integers.
    2. Hence show that \(x ^ { 2 } + 2 x + 5\) is always positive.
  1. A curve has equation \(y = x ^ { 2 } + 2 x + 5\).
    1. Write down the coordinates of the minimum point of the curve.
    2. Sketch the curve, showing the value of the intercept on the \(y\)-axis.
  2. Describe the geometrical transformation that maps the graph of \(y = x ^ { 2 }\) onto the graph of \(y = x ^ { 2 } + 2 x + 5\).

AnswerMarks Guidance
4(a)(i)\((x+1)^2 + 4\) B1 B1
4(a)(ii)\((x+1)^2 \geq 0 \Rightarrow (x+1)^2 + 4 > 0\) E1
\((\Rightarrow x^2 + 2x + 5 > 0\) for all values of \(x)\)
4(b)(i)\(x = -1\) or \(y = 4\) M1
Minimum point is \((-1, 4)\)A1
4(b)(ii)[Sketch shows parabola opening upward with vertex marked] B1
y-intercept 5 or \((0, 5)\) marked or statedB1 y-intercept 5 or (0, 5) marked or stated
4(c)Translation (not shift, move etc) through \(\begin{pmatrix} -1 \\ 4 \end{pmatrix}\) (or 1 left, 4 up etc) E1 M1 A1
Total: 10 marks
**4(a)(i)** | $(x+1)^2 + 4$ | B1 B1 | $p = 1$ and $q = 4$

**4(a)(ii)** | $(x+1)^2 \geq 0 \Rightarrow (x+1)^2 + 4 > 0$ | E1 | Condone if they say $(x+1)^2$ positive and adding 4 so always positive
| $(\Rightarrow x^2 + 2x + 5 > 0$ for all values of $x)$ |  |  |

**4(b)(i)** | $x = -1$ or $y = 4$ | M1 | ft their $x = -p$ or $y = q$
| Minimum point is $(-1, 4)$ | A1 |

**4(b)(ii)** | [Sketch shows parabola opening upward with vertex marked] | B1 | Sketch roughly as shown
| y-intercept 5 or $(0, 5)$ marked or stated | B1 | y-intercept 5 or (0, 5) marked or stated

**4(c)** | Translation (not shift, move etc) through $\begin{pmatrix} -1 \\ 4 \end{pmatrix}$ (or 1 left, 4 up etc) | E1 M1 A1 | E1: and NO other transformation stated. M1: either component correct or ft their $-p, q$. A1: correct translation (M1, A1 independent of E mark)

**Total: 10 marks**

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4
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Express $x ^ { 2 } + 2 x + 5$ in the form $( x + p ) ^ { 2 } + q$, where $p$ and $q$ are integers.
\item Hence show that $x ^ { 2 } + 2 x + 5$ is always positive.
\end{enumerate}\item A curve has equation $y = x ^ { 2 } + 2 x + 5$.
\begin{enumerate}[label=(\roman*)]
\item Write down the coordinates of the minimum point of the curve.
\item Sketch the curve, showing the value of the intercept on the $y$-axis.
\end{enumerate}\item Describe the geometrical transformation that maps the graph of $y = x ^ { 2 }$ onto the graph of $y = x ^ { 2 } + 2 x + 5$.
\end{enumerate}

\hfill \mbox{\textit{AQA C1 2009 Q4 [10]}}