Vertex of parabola is (0,40)

The helicopter must be above 40.

This real is simply a real world application of how to find the vertex of a parabola

**Practice** Problems

The ball lands at the solution of this quadratic equation. There are two solutions. One at 2 and the other at − 2. This picture assumes that Joseph threw the ball to the right so that the whiffle balls lands at 2

You can solve this quadratic by factoring or by using the quadratic formula

The ball lands at the solution of this quadratic equation. There are two solutions. One at 2 and the other at − 2. This picture assumes that Joseph threw the ball to the right so that the whiffle balls lands at 1.

The ball hits the ground at d = 0. To find the value of t at this point we must solve this quadratic equation.

0 = −t² + 36

t² = 36

t = 6

We want to find when d=0, which represents the moment when the ball hits the ground.

d = 0, when 0 = -5t² +60

5t² = 60

t² = 60 ÷5 = 12

$$ \sqrt{12} \approx 3.5 $$

Identify all of the occurrences of 't' and substitute the input in

$$ h( \color{blue}{t }) = -5\color{blue}{t}^2 + 40 \color{blue}{t} + 1.2 \\ h( \color{blue}{4 }) = -5 \cdot \color{blue}{4}^2 + 40 \cdot \color{blue}{4} + 1.2 $$

Compute result

$$ h( \color{blue}{4 }) = -5 \cdot \color{blue}{4}^2 + 40 \cdot \color{blue}{4} + 1.2 \\ h( \color{blue}{4}) = \color{red}{81.2} $$

$$ h(\color{blue} {input } ) =\color{red} {output} \\ h(\color{blue} {4} ) =\color{red} {81.2} $$

Here is a picture of graph of projectile's path with the point $$ (\color{blue} {t}, \color{red} {h(t)}) (\color{blue} {4} , \color{red} {81.2}) $$<:>

Graph generated by Meta Calculator's graphing calc