﻿ Parabolas: Equation of, Characteristics of, and Graphs of Parabolas

# Parabola in Real Life

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

### 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

(Note: t = -6 is also a solution of this equation. However, only the positive solution is valid since we are measuring seconds.)

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

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

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 $$(\color{blue} {t}, \color{red} {h(t)}) (\color{blue} {4} , \color{red} {81.2})$$<:>

Graph generated by Meta Calculator's graphing calc

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