TOTAL 22 marks
- The plots shows the curve $y=x^2+4x-1$
- Use the formula to solve the equation $x^2+4x-1=0$ giving your answers to 2 d.p.\[ (x+2)^2 - 5 = 0 \\ x = -2 \pm \sqrt{5} \\ x=0.24 \text{ or } -4.24\]
- $x^2+4x-1=k$ has no solutions, where k is an integer. Use the plot above to find the maximum value of k \[k = -6\]
[4 marks]
- Complete the square $x^2+8x-4$ and use your answer to solve \[x^2+8x-4=0 \\ (x+4)^2 - 20 = 0 \\ x=-4 \pm \sqrt{20} \\ x=-4 \pm 2\sqrt{5}\]
[3 marks]
- Y varies indirectly as the square root of x. If y=7 when x=4, then:
- Find a formula for y in terms of x \[y=\frac{k}{\sqrt x} \\ k = 7 \times 2=14 \\ \therefore y=\frac{14}{\sqrt x} \]
- Find y when x=10 to 3 s.f. \[ y = 14 \div \sqrt{10} = 4.43 \]
- Find x when y=2.2 \[ x=(\frac{14}{y})^2 = (\frac{14}{2.2})^2 = 40.5 \]
[4 marks]
- Find the image of the point (-2, -3) under the transformation represented by M
\[
\begin{pmatrix}
4 & -3 \\
-5 & -7
\end{pmatrix}
\begin{pmatrix}
-2 \\
-3
\end{pmatrix}
=
\begin{pmatrix}
1 \\
31
\end{pmatrix}
\\ \text{Point is } (1,31) \]
[2 marks]
- Find the 2x2 matrix which represents a reflection in the x axis.
\[
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
\]
[2 marks]
- Write $2x^2+6x -3$ in the form $a(x+b)^2 +c$. \[2(x+\frac{3}{2})^2 - \frac{15}{2} \]
[3 marks]
- Draw up a table of values for $-2 \leq x \leq 3$ for the function $y=x^3 -2x^2+x-3$ and use it to sketch the curve neatly (and with appropriate scales).
\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
\ x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline
\ y & -21 & -7 & -3 & -3 & -1 & 9 \\ \hline
\end{array}
\]
[4 marks]
