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Visual patterns

May 15, 2019

we’re practicing how to make equations from visual patterns in grade 10. We are using colours to show different terms. In this we have 2 n by n squares, and 3 groups of n in each figure. We see figure 1,2,3 shown.we can make tables of values and show that our pattern is growing quadratically, since the second differences are the same.sometimes there are a few ways to view a pattern, and we can make different equations (that would all simplify to the same thing in the end)we looked at how to find the vertex of this equation. We decided to complete the square just to see if we could, even though the numbers are not so friendly. We use the same visual method, and use fractions or decimals.

We could also have used the quadratic formula to find the roots, then found the axis of symmetry and then the vertex. It’s great there are so many options!

Eqao practice stations

May 15, 2019

We are working hard to get ready for our linear relations test and also practice EQAO questions at the same time.

We are getting better at using our reasoning to work through multiple choice problems

And we are showing our thinking in many different ways.

Word problems

May 14, 2019

Grade 10s are working on quadratic word problems. We’re looking at maximum areas for fields, and maximum revenue for bushels of potatoes. In any situation we need to first identify variables, make an equation, or two, with the goal to make something quadratic. We are looking here for the maximum area or revenue, so we want to find the vertex (sommet) so we need to remember all the ways to do that, and hopefully choose a way that is quick. It’s often easy to find the zeros, the midpoint, and then the vertex.our revenue problem involved a farmer who started with 600 bushels of potatoes to be sold at 2$/ bushel. If she waits, she gets 100 more bushels a week, but the price drops by 10 cents a bushel. She needs to know when to harvest, to get the highest revenue.

We can make a table and solve, but there are sometimes situations where a table becomes a very long tedious method. From the table we notice it is quadratic data. We notice that revenue is equal to price times number of items sold, and we also can make expressions for price and number based on waiting x number of weeks.using the table we see the maximum revenue occurs at 7 weeks. We also see the y intercept is 1200$. If we expand to make the trinomial we see the constant is 1200. We also noticed something neat. The 2nd differences are always double the “a” value of our quadratic. We can use the trinomial form and the quadratic formula, or go back to the factored form and solve for the zeros, we could also complete the square..so many ways to get to the vertex! We find the ridiculous cases (the “stupid” case) where revenue is equal to zero. In this question if the farmer waits 20 weeks there will be lots of potatoes to sell but the price will have dropped to zero. If we sold 6 weeks ago the price would have been higher, but there would have been no potatoes to sell! we look to the graph. We can plot those ridiculous cases with 0 revenue, and then find the sweet spot in the middle where the revenue is maximized. We calculate the midpoint to be 7 and then can calculate the revenue when x is 7 by using our formula.

It’s a bit of a surprise for some of us to be seeing parabolas appear in problems that do not involve kicking a ball. We are seeing them in area problems, price problems, and patterning too. We follow the same steps as we’ve done before, but now need to keep a context in mind and make good choices of the path through the problem.

Speed dating

May 14, 2019

Grade 9s are working on graphing lines, and making equations for graphs.

We’re working across the table from eachother in “speed dating” style. Both partners solve the problem together, then one side stays as expert, the other side rotates and tries a new question. The first side helps their new partner to become an expert before they themselves shuffle to the next question.

There’s a lot of reciprocal teaching happening, lots of clarifying questions being asked, and some big “aha” moments!

We looked all together at how to calculate the constant (initial value/y intercept) when we have a line that passes through 2 points.

by graphing we could get close. We know the slope is 2/10 or 1/5 and we know that the y intercept is just less than 2.

We can do better than that! Some of us figured that for each 1 square we move forward, we should go up by 0.2 since that’s 1/5. Using that logic the y intercept is 1.8

We looked at how a table can be helpful to find the slope as well. And how this connects to the equation for slope.

It’s all “rise over run” no matter how you look at it.

And the slope ends up being 1/5 again!

We can calculate the y intercept (b) by subbing in a point on the line for x and y. We then use algebra to get b.

it doesnt matter what point we choose to use, b will be the same! 1.8 in this case.

If we are not keen on algebra, we looked at how desmos sliders can be helpful, and also how desmos can do regression for us! Here’s a different equation, of a line that goes through the points that are in the table above. Technology is pretty cool!

All the forms of quadratics

May 11, 2019

Some quadratics work nicely in all forms. Here’s an example of one that can be factored, and also one that we can complete the square of, with no decimals. This is not always the case. Some cannot be factored, some have no roots. Some have a vertex (sommet) that has decimals. We can always complete the square, but it can be a lot messier than in this case.

Beans!

May 10, 2019

We have our first flower.

Multiple representations

May 10, 2019

Here’s an example of all the ways we can represent a linear relation. A visual pattern, an equation, a graph, vocabulary, and a table of values. The rate is shown in black, and the constant in green.

We worked on solving a problem in class too about a fridge repair. The first repair job cost 176$ and took 2 hours. The next job took 3.5 hours and cost 263$. We made a graph and found the rate (using the triangle). The price increased by 87$ for 1.5 hours of increased time. We looked for the unit rate which is 58$/h. The constant is found on the vertical axis, but we can’t know exactly what it is without some more calculations

we can use our table of values and count backwards, for every hour we back track, the cost decreases by 58$. The constant is 60$.

We can also make an equation with the cost and time. “B” is our constant, which we are trying to calculate. We substitute values that we know. For time we used 2 hours and for cost we used 176$. We already know the hourly rate is 58$/h. We can then calculate the value of “b” which is the same as we got from our table.