Question

Using a certain ramp at the skate 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 horizontal distance \( x \) from the jumping platform at which the skateboarder is at their minimum height, we need to determine the vertex of the given quadratic equation:

\[
y = 0.25x^2 - 4x + 19
\]

For a quadratic equation in the form \( ax^2 + bx + c \), the x-coordinate of the vertex (which gives the minimum or maximum point) can be found using the formula:

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

In our equation, \( a = 0.25 \) and \( b = -4 \). We can substitute these values into the vertex formula:

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

Calculating inside the equation:

\[
x = \frac{4}{0.5} = 8
\]

Thus, the skateboarder is **8 feet** from the jumping platform when they are at their minimum height off the ground.