Using a certain ramp at the state park, a skater's height off the ground, in feet, is modeled by the equation y=0.25x^2-4x+19, where x is the skater's horizontal distance from the jumping platform. How far from the jumping platform is a skateboarder when they are at their minimum height off the ground?

Answer 1

To find the distance from the jumping platform when the skater is at their minimum height, we need to identify the vertex of the parabola described by the equation \( y = 0.25x^2 - 4x + 19 \). The vertex of a quadratic equation in the form \( y = ax^2 + bx + c \) occurs at the \( x \)-coordinate given by the formula:

\[ x = -\frac{b}{2a} \]

In our equation, we have:

• \( a = 0.25 \)
• \( b = -4 \)

Now we can substitute these values into the vertex formula:

\[ x = -\frac{-4}{2 \times 0.25} = \frac{4}{0.5} = 8 \]

Thus, the distance from the jumping platform when the skateboarder is at their minimum height off the ground is:

\[ \boxed{8} \text{ feet} \]