The problem asks to find the coordinates of the turning point of the curve given by the quadratic equation $y = x^2 + ax + b$. We know that:

• The points $(0, -5)$ and $(5, 0)$ lie on the curve.
• $a$ and $b$ are integers.

Step 1: Substituting the points into the equation

Using the point $(0, -5)$:

$y = x^2 + ax + b \quad \text{at} \, (0, -5):$ $-5 = 0^2 + a(0) + b \implies b = -5$

Using the point $(5, 0)$:

$y = x^2 + ax + b \quad \text{at} \, (5, 0):$ $0 = 5^2 + a(5) - 5 \implies 0 = 25 + 5a - 5 \implies 5a = -20 \implies a = -4$

Thus, the equation of the curve becomes:

$y = x^2 - 4x - 5$

Step 2: Finding the turning point

The turning point (vertex) of a parabola $y = x^2 + ax + b$ occurs at $x = -\frac{a}{2}$. In this case:

$x = -\frac{-4}{2} = 2$

To find the corresponding $y$-coordinate, substitute $x = 2$ into the equation:

$y = 2^2 - 4(2) - 5 = 4 - 8 - 5 = -9$

Final Answer:

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

Would you like more details on any step or have any further questions?

Here are 5 follow-up questions to extend your understanding:

1. How can you verify the turning point of a quadratic equation using calculus?
2. What is the general form of the equation for the vertex of any parabola $y = ax^2 + bx + c$?
3. How does the value of $a$ in the equation affect the shape and direction of the parabola?
4. How do you determine whether the turning point is a maximum or minimum for a quadratic function?
5. What role does the discriminant play in determining the number of real roots of a quadratic equation?

Tip: The turning point of a quadratic function can be easily found using the formula $x = -\frac{b}{2a}$ for the vertex, where $b$ and $a$ are coefficients from the general form $ax^2 + bx + c$.