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Grade 12s were looking at what equation would be needed to create this graph. Good luck for the test tomorrow grade 12s!

Grade 9s looked at their bean graph to try to determine the growth rate for kidney beans and black eyed peas.

We noticed that the red beans grew earlier, and faster than the black eyed peas. This meant that their points made a steeper line of best fit, and their x intercept was lower than that of the black eyed peas.

We saw that sometimes for an individual bean the trend was not linear. More often than not the bean started at a higher rate, and then grew less and less each day after a certain point. This trend is non-linear. We did not have any magic beans that got increasingly tall each day by a bigger amount.

We looked at how to find the growth rate of a bean between 2 days. We calculate the number of cm grown and divide that by the number of days elapsed between measurements.

For the group of beans we can do this for our line of best fit.

Some groups found many averages

We looked at how to do a regression on desmos. We can make desmos do the figuring for us, and come up with the line of best fit. We can use the line created by desmos to do interpolation and extrapolation. We can also notice the intersection point. At that time the two plants were the same height (on average).

Technology can save us a lot of time!

We looked at question 7 together. The steps we use:

1. Make a picture

2. Introduce variables and make an equation or maybe more than one

3. Trinomial=0

4. Factor

5. Parentheses=0

6. Solve

7. Concluding statement

Grade 9s are working on equation solving today.

Here’s an example of a solution. We tried some pretty complex problems, involving lots of simplification, and even fractions.

We looked at how to divide fractions, or how to multiply by the denominator to eliminate the fractions…

We also looked at what fractions mean, and how we can represent them

We looked at what a negative fraction means, and realized that it doesn’t really matter where the negative sign is within the fraction.

We’ve got a quiz tomorrow to check our understanding.

Solving equations with algebra tiles: we first represent what’s on both sides of the equal sign using algebra tiles. There is an imaginary wall between the sides (where the equal sign is)

We next add 2x to both sides (to create 0x on the right)

Next we simplify (eliminate all zero pairs)

Next we eliminate the constant that’s on the same side as the x term (we place 4 blue squares on each side)

Next we simplify. We know that -x=2 so we can deduce (or divide both sides by the coefficient) that x=-2

What do you notice?  What do you wonder?

What would happen if the red function were the numerator and the blue function were the denominator?  What do you know about the quotient function?

What would happen if the blue function were the numerator and the red function were the denominator?  What do you know about the quotient function?

We had a look at the problem in class today. We realized that we don’t need the numbers on the graph, or the equation even, to get an idea of a rational function’s graph.

We looked for places where the denominator function is zero. That’s where we’d find vertical asymptotes. We looked for where the numerator function is zero, and that’s where the zeros of the rational function will be. We looked at the signs, and relative values of the numerator and denominator and strategically looked to each side of the asymptotes, and at the end behaviours.

some groups had time to look at the rational function of “blue” divided by “red” as well. The function looks very different at the ends. We know that when the degree of the numerator is less than the degree of the denominator, the function will have a horizontal asymptote at 0. Above, the numerator had a degree that was 2 greater than the denominator, which led to a parabolic asymptote.

We’re making connections between equations and graphs of quadratics.

We are able to graph parabolas in factored form. We know how to determine equations from graphs too. It was a tasty parabola! Y=2(x-1)(x-3)

Today in grade 9 we made progress in our equation solving skills.

We are sometimes solving by inspection, and then showing a full verification to prove that our answer is correct.

When we can’t solve by inspection, we need to have a strategy. We are able to maintain the balance if we do the same thing on both sides of the equal sign.

We are showing our steps, simplifying, and then isolating the variable. We are certainly building on our algebra skills!