1. Problem statement: Triangle A'B'C' is the image under a reflection (diagram not provided), and the tasks are to draw the mirror line and determine its equation in the form $y=mx+c$, to describe a single transformation mapping triangle A'B'C' onto triangle A''B''C'', and for the cubic $y=x^3+5x^2+Px-18$ with gradient $-15$ at $x=-4$ to find $P$, to find the equation of the normal at $x=-1$, and to find the turning points. 2. Mirror line (what we can and cannot do): Because the diagram and coordinates are not provided I cannot draw the mirror line or give its numeric equation without the coordinates of corresponding points A, B, C and their images. 3. Mirror line (method to construct and its equation): For each corresponding point pair, e.g. $A$ and $A'$, compute the midpoint $M$ by $M((x_A+x_{A'})/2,(y_A+y_{A'})/2)$. 4. Mirror line (method continued): Compute the slope of the segment $AA'$ by $m_{AA'}=(y_{A'}-y_A)/(x_{A'}-x_A)$. 5. Mirror line (final formula): The mirror line is perpendicular to $AA'$ so its slope is $m=-1/m_{AA'}$ provided $m_{AA'}\neq 0$. 6. Mirror line (line equation): Using the midpoint $M(x_M,y_M)$ the mirror line has equation $y=mx+c$ with $c=y_M-mx_M$. 7. Single transformation mapping A'B'C' onto A''B''C'' (what we can and cannot do): Without the diagrams or coordinates I cannot state the specific single transformation numerically, but here is how to identify it in practice. 8. Identify translation: If every vertex is shifted by the same vector $\vec{v}=(\Delta x,\Delta y)$ then the single transformation is the translation $T_{\vec{v}}$ where $\Delta x=x_{A''}-x_{A'}$ and $\Delta y=y_{A''}-y_{A'}$. 9. Identify rotation: If distances from a common point $O$ are preserved and orientation is preserved but points rotate by the same angle, then the single transformation is a rotation about $O$ by angle $\theta$ found from the vector directions. 10. Identify reflection: If each midpoint of a segment joining a point and its image lies on the same straight line and orientation is reversed, then the single transformation is a reflection in that common line (the mirror line found above). 11. Identify enlargement: If corresponding lengths are in a constant ratio $k\neq 1$ from a common center, then the single transformation is an enlargement (scaling) about that center by factor $k$. 12. Summary for geometry parts: Provide coordinates or the diagram to obtain a unique numeric equation for the mirror line and to state the exact single transformation. 13. Cubic: start by differentiating the cubic $y=x^3+5x^2+Px-18$ to get the gradient function $y'=3x^2+10x+P$. 14. Find $P$ using the condition $y'(-4)=-15$: evaluate $y'(-4)=3(-4)^2+10(-4)+P=48-40+P=8+P$. 15. Solve $8+P=-15$ to get $P=-23$. 16. Equation of the normal at $x=-1$: first compute the tangent gradient at $x=-1$ using $P=-23$, giving $y'(-1)=3(1)+10(-1)+(-23)=3-10-23=-30$. 17. The normal gradient is the negative reciprocal of the tangent gradient, so $m_{\text{normal}}=1/30$. 18. Find the point on the curve at $x=-1$: $y(-1)=(-1)^3+5(-1)^2+(-23)(-1)-18=-1+5+23-18=9$. 19. The normal line through $(-1,9)$ with slope $1/30$ has point-slope form $y-9=(1/30)(x+1)$, which simplifies to $y=(1/30)x+271/30$. 20. Turning points: set $y'=0$ with $P=-23$ to solve $3x^2+10x-23=0$. 21. Compute the discriminant $D=10^2-4\cdot 3\cdot(-23)=100+276=376$. 22. Solve for $x$: $x=(-10\pm(376)^{1/2})/6$, which simplifies to $x=(-5\pm(94)^{1/2})/3$. 23. Numerical approximations: $(94)^{1/2}\approx 9.6953597148$, so the turning $x$-coordinates are $x_1\approx(-5+9.6953597148)/3\approx 1.5651199049$ and $x_2\approx(-5-9.6953597148)/3\approx -4.8984532383$. 24. Evaluate $y$ at these $x$-values (using $P=-23$) to get approximate turning points: for $x_1\approx 1.5651199049$ we find $y\approx -37.9161$ and for $x_2\approx -4.8984532383$ we find $y\approx 97.1010$. 25. Final answers: $P=-23$. 26. Final answers: equation of the normal at $x=-1$ is $y=(1/30)x+271/30$. 27. Final answers: turning points at $\left(\dfrac{-5+(94)^{1/2}}{3},\,\approx -37.9161\right)$ and $\left(\dfrac{-5-(94)^{1/2}}{3},\,\approx 97.1010\right)$.