Question

a diver jumps into a pool from a board ten feet above the surface of the water. her height, h, at any time t seconds after jumping is given by the equation ( h = t^{2}-8t + 10 ). what is the minimum height that she reaches? (note: negative values of h mean the diver is below the water.)
-26 feet
-10 feet
-6 feet
-4 feet
to the nearest centimeter, chandra is 158 centimeters tall. what is the smallest possible value for chandras actual height?
157.5 centimeters
157 centimeters
158.4 centimeters
158 centimeters

Explanation:

Step1: Find the vertex of the quadratic function

For a quadratic function \(y = ax^{2}+bx + c\) (here \(h=t^{2}-8t + 10\), so \(a = 1\), \(b=-8\), \(c = 10\)), the \(t\) - coordinate of the vertex is given by \(t=-\frac{b}{2a}\).

$$t=-\frac{-8}{2\times1}=4$$

Step2: Substitute \(t = 4\) into the height function

Substitute \(t = 4\) into \(h=t^{2}-8t + 10\)

$$h=(4)^{2}-8\times4 + 10=16-32 + 10=-6$$

For the second question:
When a measurement is given to the nearest centimeter, if the measured value is \(x\) (here \(x = 158\)), the smallest possible value is \(x-0.5\) (using the concept of rounding). So the smallest possible value for Chandra's actual height is \(158 - 0.5=157.5\) centimeters.

Answer:

For the first question: \(-6\) feet.
For the second question: \(157.5\) centimeters.