Experiment with cases and illustrate an explanation of the effects on the graph using technology.

Include recognizing even and odd functions from their graphs and algebraic expressions for them.

A border of flowers are planted around the garden so that the area of the flowers is exactly one half the area of the field. The equation area of the garden will be Area = (30 – 2x)(40 – 2x) Area needs to be ½ the area of the field so will be 600m2 600 = (30 – 2x)(40 – 2x) 600 = 1200 – 140x – 4x2 4x2 140x – 600 = 0 (divide by 4) x2 35x – 150 = 0 (factor) (x 30)(x 5) = 0 so x = -30 or x= 5 since distance is positive x = 5.

Give students problems that allow for multiple points of entry and let them use lots of time and technology to develop their own methods to solve them--after several days, you will be excited to see all the different methods they come up with!

All quadratic equations can be put in standard form, and any equation that can be put in standard form is a quadratic equation.

## Solving Problems With Quadratic Functions Problem Solving Resume

In other words, the standard form represents all quadratic equations.

MPM2D Unit 3 – Quadratic Applications Solving problems with Quadratic Relations. Type II - Projectile Motion problems where you are given the zeros Jon fires a toy rocket into the air from the ground. Substitute in x = 5 y = -1.25(5)(5- 10) y = 31.25 Therefore the greatest height is 31.25 m MPM2D Unit 3 – Quadratic Applications Type III - Revenue Problems A restaurant determines that each 10 cent increase in the price of a salad results in 25 fewer salads being sold.

Sample types of problems Type I - Projectile Motion problems where you are given the equation. The rocket is in the air for 10 s before it falls to the ground. The usual price for a salad is .00 and the restaurant sells 300 salads each day.

So we need to figure out at which times does h equal 0. And so we have 8t squared minus 10t minus 25 is equal to 0. And we could complete this square here, or we can just apply the quadratic formula, which is derived from completing the square. Well, that's just positive 100, minus 4 times a, which is 8, times c, which is negative 25. Now, we have to remember, we're trying to find a time.

So we're really solving the equation 0 is equal to negative 16t squared plus 20t plus 50. Or if you're comfortable with this on the left hand side, we can put on the left hand side. And so a time, at least in this problem that we're dealing with, we should only think about positive times.

## Comments Solving Problems With Quadratic Functions

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