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Let's solve question 12 from the image.

### Given Information:
- The path of the ball is modeled by the equation \( y = h + kx - \frac{1}{2}x^2 \).
- The ball passes through a hoop at the point \( x = 2 \) and \( y = 5 \).
- We are tasked to find the values of \( h \) and \( k \), and the value of \( x \) at which the ball hits the ground (i.e., when \( y = 0 \)).

### Step 1: Use the point \( (2, 5) \) to find an equation.
Substitute \( x = 2 \) and \( y = 5 \) into the equation \( y = h + kx - \frac{1}{2}x^2 \).

\[
5 = h + k(2) - \frac{1}{2}(2^2)
\]
Simplifying:

\[
5 = h + 2k - \frac{1}{2}(4)
\]
\[
5 = h + 2k - 2
\]
\[
7 = h + 2k \quad \text{(Equation 1)}
\]

### Step 2: Use the fact that the ball is at its highest point when \( x = 0 \) (the vertex of the parabola).
At the highest point, \( x = 0 \) and the derivative of \( y \) with respect to \( x \) is zero. First, find the derivative:

\[
\frac{dy}{dx} = k - x
\]
At the highest point, set \( x = 0 \):

\[
k = 0
\]
So, the slope of the curve at the highest point (where the ball is at its peak) is 0, which means the constant \( k \) represents the initial velocity.

### Step 3: Find the value of \( x \) when the ball hits the ground (i.e., \( y = 0 \)).


Amy throws a ball so that when it is at its highest point, it passes through a hoop. The path of the ball is modelled by the equation y = h + kx − ½x², where y is the height of the ball above the ground and x is the horizontal distance from the point at which the ball was thrown. The center of the hoop is at the point where x = 2 and y = 5. Find the values of h and k, and the value of x at which the ball hits the ground.

Learn how to solve a parabola-based question where a ball passes through a hoop, using the quadratic equation to find values for h, k, and the ball's horizontal distance when it hits the ground. Suitable for Grades 10-12.

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