The problem presented is about finding the turning point of a quadratic curve. Let's break it down.

Problem details:

• The equation of the curve is given by $y = x^2 + ax + b$, where $a$ and $b$ are integers.
• Two points on the curve are given: $(0, -5)$ and $(5, 0)$.
• We are asked to find the coordinates of the turning point of the curve.

Steps to solve:

1. Substitute the points into the equation to find $a$ and $b$:

   For $(0, -5)$, substitute $x = 0$ and $y = -5$ into the equation: $-5 = 0^2 + a(0) + b \implies b = -5.$

   Now, for $(5, 0)$, substitute $x = 5$ and $y = 0$: $0 = 5^2 + a(5) - 5 \implies 0 = 25 + 5a - 5 \implies 5a = -20 \implies a = -4.$

2. Rewrite the equation of the curve: The equation now becomes: $y = x^2 - 4x - 5.$

3. Find the turning point: The turning point of a quadratic function $y = x^2 + ax + b$ occurs at $x = -\frac{a}{2}$.

   Here, $a = -4$, so: $x = -\frac{-4}{2} = 2.$

   To find the $y$-coordinate of the turning point, substitute $x = 2$ into the equation $y = x^2 - 4x - 5$: $y = 2^2 - 4(2) - 5 = 4 - 8 - 5 = -9.$

Final Answer:

The turning point of the curve is at $(2, -9)$.

Would you like more details on this, or do you have any questions?

Here are 5 related questions you might want to explore:

1. How do we determine the axis of symmetry for a quadratic equation?
2. What is the significance of the discriminant in quadratic equations?
3. How would the curve shift if the values of $a$ or $b$ changed?
4. How do you find the roots of the quadratic equation $y = x^2 - 4x - 5$?
5. What is the general form of the vertex form of a quadratic equation?

Tip: The vertex (or turning point) of any quadratic equation $y = ax^2 + bx + c$ can also be found by completing the square!