1. Problem statement: Triangle A'B'C' is given as the image under a reflection (diagram unspecified); tasks are: draw the mirror line and determine its equation in the form $y = mx + c$, describe a single transformation mapping triangle A'B'C' onto triangle A''B''C'', and for the cubic $y = x^3 + 5x^2 + P x - 18$ with gradient $-15$ at $x = -4$ find $P$, the equation of the normal at $x = -1$, and the turning points. 2. (Reflection drawing) The mirror line is the perpendicular bisector of the segment joining any point to its reflected image, so to draw it: join $A'$ to $A''$ and find the midpoint $M_A$, draw the line perpendicular to $A'A''$ through $M_A$, and check that the same line is the perpendicular bisector of $B'B''$ and $C'C''$; if it is, that line is the mirror line. 3. (Reflection equation method) Let corresponding points be $A'(x_1,y_1)$ and $A''(x_2,y_2)$ in coordinate form, then the midpoint is $M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ and the slope of $A'A''$ is $m_{A'A''}=\frac{y_2-y_1}{x_2-x_1}$, so the slope of the mirror line is $m=-\frac{x_2-x_1}{y_2-y_1}$ because it is perpendicular; the mirror-line equation through the midpoint is $y-\frac{y_1+y_2}{2}=m\left(x-\frac{x_1+x_2}{2}\right)$ and rearranges to $y=mx+c$ with $c=\frac{y_1+y_2}{2}-m\frac{x_1+x_2}{2}$. 4. (Reflection final description) Therefore the single transformation mapping triangle A'B'C' onto triangle A''B''C'' is a reflection in the mirror line found above. 5. (Calculus: derivative and P) Differentiate the cubic to get $\frac{dy}{dx}=3x^2+10x+P$. 6. Evaluate the gradient condition at $x = -4$: $3(-4)^2+10(-4)+P=-15$. 7. Compute: $3\times16-40+P=-15$ so $48-40+P=-15$, hence $8+P=-15$ and therefore $P=-23$. 8. (Normal at $x = -1$) First compute the tangent gradient at $x = -1$ using $P=-23$: $\frac{dy}{dx}\bigg|_{x=-1}=3(-1)^2+10(-1)+P=3-10+P=-7+P=-7-23=-30$. 9. The normal gradient is the negative reciprocal, $m_{\text{normal}}=\frac{1}{30}$. 10. Find the point on the curve at $x = -1$: $y=(-1)^3+5(-1)^2+P(-1)-18=-1+5- P -18=-14 - P$ and with $P=-23$ this gives $y=-14+23=9$, so the point is $(-1,9)$. 11. Equation of the normal through $(-1,9)$ is $y-9=\frac{1}{30}(x+1)$, which rearranges to $y=\frac{1}{30}x+\frac{271}{30}$. 12. (Turning points) Turning points occur where $\frac{dy}{dx}=0$, so solve $3x^2+10x+P=0$ and substitute $P=-23$ to get $3x^2+10x-23=0$. 13. Compute the discriminant: $D=10^2-4\times3\times(-23)=100+276=376$ and $\sqrt{376}=2\sqrt{94}$, so the stationary $x$-values are $x=\frac{-10\pm2\sqrt{94}}{6}=\frac{-5\pm\sqrt{94}}{3}$. 14. To find the corresponding $y$-values, use the cubic and eliminate $x^3$ by using $3x^2=-10x+23$ (from the derivative equation): start with $y=x^3+5x^2-23x-18$, substitute $x^3=x\cdot x^2= x\cdot\frac{-10x+23}{3}=\frac{-10x^2+23x}{3}$, simplify to get $3y=5x^2-46x-54$, substitute $x^2=\frac{-10x+23}{3}$ and simplify to obtain $y=\frac{799\mp188\sqrt{94}}{27}$ corresponding respectively to $x=\frac{-5\pm\sqrt{94}}{3}$. 15. Therefore the turning points are exactly $\left(\dfrac{-5+\sqrt{94}}{3},\dfrac{799-188\sqrt{94}}{27}\right)$ and $\left(\dfrac{-5-\sqrt{94}}{3},\dfrac{799+188\sqrt{94}}{27}\right)$, which are approximately $(1.56512,-37.8976)$ and $(-4.89846,97.0824)$ respectively. 16. Final answers summary: $P=-23$. 17. Final answers summary: Normal at $x=-1$ is $y=\dfrac{1}{30}x+\dfrac{271}{30}$. 18. Final answers summary: Turning points as given in step 15. 19. Final answers summary: The single transformation mapping A'B'C' onto A''B''C'' is a reflection in the mirror line whose equation is given by the perpendicular-bisector formula in step 3 (substitute the diagram coordinates to obtain numeric $m$ and $c$).