OCR PURE — Question 5 5 marks

Exam BoardOCR
ModulePURE
Marks5
PaperDownload PDF ↗
TopicSolving quadratics and applications
TypeFinding quadratic constants from algebraic conditions
DifficultyModerate -0.8 This question tests basic understanding of vertex form of a quadratic and simple transformations. Part (a) requires recognizing that the vertex form y = a(x+b)² + c has its stationary point at (-b, c), making b=3 and c=2 immediate. Part (b) involves applying a horizontal translation and substituting a point to find a. All steps are routine with no problem-solving insight required, making this easier than average.
Spec1.02e Complete the square: quadratic polynomials and turning points1.02w Graph transformations: simple transformations of f(x)

5 A curve has equation \(y = a ( x + b ) ^ { 2 } + c\), where \(a , b\) and \(c\) are constants. The curve has a stationary point at \(( - 3,2 )\).
  1. State the values of \(b\) and \(c\). When the curve is translated by \(\binom { 4 } { 0 }\) the transformed curve passes through the point \(( 3 , - 18 )\).
  2. Determine the value of \(a\).

Question 5:
Part (a):
AnswerMarks Guidance
AnswerMarks Guidance
Stationary point at \((-3, 2) \Rightarrow b = 3,\ c = 2\)B2 (1.1, 1.1) B1 for one correct
[2 marks]
Part (b):
AnswerMarks Guidance
AnswerMarks Guidance
Translated curve is \(y = a(x + (b-4))^2 + c\)M1* Translates \(y\) by either \(\binom{\pm4}{0}\) only. Possible to translate \((3,-18)\) by \(\binom{\pm4}{0}\) and use original curve to find \(a\)
\(-18 = a(3 + (b-4))^2 + c \Rightarrow a = \ldots\)M1dep* Substitutes \((3, -18)\) into translated curve and finds a value for \(a\)
\(a = -5\)A1
[3 marks]
# Question 5:

## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Stationary point at $(-3, 2) \Rightarrow b = 3,\ c = 2$ | B2 (1.1, 1.1) | B1 for one correct |

**[2 marks]**

## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Translated curve is $y = a(x + (b-4))^2 + c$ | M1* | Translates $y$ by either $\binom{\pm4}{0}$ only. Possible to translate $(3,-18)$ by $\binom{\pm4}{0}$ and use original curve to find $a$ |
| $-18 = a(3 + (b-4))^2 + c \Rightarrow a = \ldots$ | M1dep* | Substitutes $(3, -18)$ into translated curve and finds a value for $a$ |
| $a = -5$ | A1 | |

**[3 marks]**
5 A curve has equation $y = a ( x + b ) ^ { 2 } + c$, where $a , b$ and $c$ are constants. The curve has a stationary point at $( - 3,2 )$.
\begin{enumerate}[label=(\alph*)]
\item State the values of $b$ and $c$.

When the curve is translated by $\binom { 4 } { 0 }$ the transformed curve passes through the point $( 3 , - 18 )$.
\item Determine the value of $a$.
\end{enumerate}

\hfill \mbox{\textit{OCR PURE  Q5 [5]}}