We are tasked with solving for $a$ and $b$ in the quadratic equation:

$y = x^2 + ax + b$

where the turning point is given as $(4, 5)$.

Step 1: Use the turning point condition

At the turning point, the derivative of $y$ with respect to $x$ is zero. The derivative is:

$\frac{dy}{dx} = 2x + a$

At the turning point $x = 4$, we set $\frac{dy}{dx} = 0$:

$2(4) + a = 0 \quad \Rightarrow \quad a = -8$

Step 2: Use the coordinates of the turning point in the equation

At $(4, 5)$, the curve satisfies the equation $y = x^2 + ax + b$. Substituting $x = 4$, $y = 5$, and $a = -8$ into the equation:

$5 = (4)^2 + (-8)(4) + b$

Simplify:

$5 = 16 - 32 + b$ $5 = -16 + b$ $b = 21$

Final Answer:

$a = -8 \quad \text{and} \quad b = 21$

Would you like a detailed explanation of any of the steps?

Related Questions:

1. How do we find the turning point of a general quadratic equation in vertex form?
2. What is the significance of the derivative being zero at the turning point?
3. Can this problem be solved graphically instead of algebraically?
4. How does the value of $a$ affect the shape and direction of the parabola?
5. What are other ways to rewrite a quadratic equation to easily identify key features?

Tip: The vertex of a parabola in the form $y = ax^2 + bx + c$ can be directly found using $x = -\frac{b}{2a}$.