##### Problem 1

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

Real world applications and problems

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

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 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

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

Step 1

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

$$ h( {\color{blue} t} ) = -5 {\color{blue} t} ^2 + 41 {\color{blue}t} + 1.2 \\ h( {\color{blue}4} ) = -5 \cdot {\color{blue}4}^2 + 40 \cdot {\color{blue} 4} + 1.2 $$

Step 2

Compute result

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

Step 3

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

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