Lesson video

Hello, Mr. Robson here, welcome to Maths.

Good choice to join me.

Today we're drawing quadratic graphs and they are cool, so let's not hang around here.

Let's get on with it.

Our learning outcome is I'll be able to generate coordinate pairs for a quadratic graph from its equation and then draw the graph.

Keywords that we'll come across today, quadratic.

A quadratic is an equation, graph, or sequence whereby the highest exponent of the variable is two.

The general form for a quadratic is axe² + bx + c.

You're gonna see a lot of those today.

We'll also come across the word, parabola.

What a lovely word that is.

A parabola is a curve where any point on the curve is an equal distance from a fixed point, which we call the focus, and a fixed straight line, the directrix.

The line of symmetry goes through the focus at right angles to the directrix.

You're gonna see a lot of parabolas today and a lot of symmetry.

Two parts to our learning, we're gonna begin by plotting quadratics.

We can plot the graph of a quadratic equation using a table of values.

We're gonna use this table to calculate the coordinate pairs for y = x².

We use substitution to find the corresponding y value for each x value, and it's sensible to start with the easiest x values.

I'm gonna start with when x = 0, 0² is 0.

I'll then move on to when x = 1, 1² is 1.

How about when x = 2? 2² is 4, when x = 3, 3² is 9, and I'll pop those values into our table.

I then need to do the negative values, <v ->1² is positive one, -2² positive four, -3² positive nine.

</v> A complete table of values.

Since 1² = -1² and 2² = -2², et cetera, we see symmetry in our table of values.

Look at those y values.

Can you see symmetry? Lovely, a pair of ones, a pair of fours, a pair of nines.

So we see this symmetry when we plot the coordinate pairs.

If I plot the coordinate pairs, -3; 9, -2; 4, -1; 1, we get this set of coordinate pairs.

I see symmetry in the table of values and I see symmetry in those coordinate pairs.

Izzy and Aisha are discussing how to draw the graph of the equation from here.

Izzy says, "I think we join these points with line segments like this".

And Izzy does that.

Aisha says, "I'm not so sure.

That means the point half a half is on your line and it shouldn't be.

A half a half there".

And Aisha is saying, "That shouldn't be on the line".

Who do you agree with? Pause, have a think, have a conversation with the people around you, I'll see you in a moment to find out.

Welcome back.

I wonder what you said.

I wonder who you agreed with.

Let's see what Izzy and Aisha figured out.

Aisha says, "Tt shouldn't be on your line.

A half a half is a coordinate pair.

When x equals a half, a half squared does not equal a half.

That does not satisfy the equation.

A half squared is a quarter", so Aisha points out the point a half a quarter should be on this line and it isn't on yours.

Izzy says, "Thanks Aisha.

Let's plot the x coordinates in steps of one half to see how the graph really looks".

"Izzy, that's an awesome idea".

When we look at this graph, y = x², and plot coordinate pairs in increments of a half, they look like that.

We can step this up into even more detail.

Increments of a quarter, increments of an eighth.

Are you starting to see the shape? If we plot every possible point that exists on this line, we get a smooth curve called a parabola.

Your quadratic graph will always form a parabola, a lovely smooth curve.

True or false? When plotting a quadratic, we join the points with straight lines.

Is that true or is it false? Once you've decided, could use one of the two statements at the bottom to justify your decision.

Pause and do that now.

Welcome back.

I do hope you went with false and justified that with, we draw a curve through the coordinate pairs to form a parabola, just like that.

How lovely.

Quadratics won't always intercept the axes at the origin.

Remember the linear graph y = x.

It looks like that.

Versus the line y = x + 3, which is there.

Can you see the difference? Adding a constant changed the y intercept but didn't change the shape of the graph.

Still talking about a straight line with a gradient of positive one, but with x + 3, we've changed the y intercept.

Will this also be true for quadratic graphs? What do you think? What's your intuition telling you? Let's find out.

Let's add a constant of three to our quadratic, instead of y = x², let's plot y = x² + 3.

It's no different.

I'll start with easiest values, x = 0, 0² + 3 is 3.

X = 1, 1² + 3 is 4, 2² + 3 is 7, 3² + 3 is 12, pop those into the table.

I'll do the negative values next.

<v ->1² that's positive one add three, that's four.

</v> <v ->2², that's positive four, add three, that's seven.

</v> <v ->3², that's positive nine, add three, that's 12.

</v> And when I populate that table, our symmetry still exists, but we no longer have 0, 0 the origin as a coordinate pair.

How will that look on a graph? Like so.

When we join a lovely smooth parabola through those coordinate pairs, you'll notice the addition of a constant hasn't changed the shape of our graph.

Still a parabola, but it has changed our y intercept.

Our y intercept is now at 0, 3.

Quick check you've got that.

I'd like to complete the table of values for this equation and also identify the y intercept.

Pause, work out those bare values now.

Welcome back.

Let's see how we did.

Hopefully you spotted that when x = 0, y = -5, when x = 1, y = -4, when x = 3, y = positive four, when x = -2, y = -1.

And when you populate your table of values it looks like so.

And we can see that the y intercept will be 0, -5.

Of course you had that as soon as you substituted in x = 0.

There's our y intercept, and that's what the graph looks like.

We've still got a parabola, but our y intercept is no longer the origin because we added a constant.

Now instead of adding a constant, let's change the coefficient of our x² term and see what happens.

Instead of graphing y = x², let's graph y = 2x².

What do you think is going to happen? You might wanna pause this video and make a few predictions, before we find out.

Let's see how this graph's going to look.

I'm going to start with x = 0.

Two lots of 0² is 0, when x = 1, two lots of 1² is 2, two lots of 2² is 8, two lots of 3² is 18, do note our knowledge of the priority of operations can be used here, or should be used here.

The x value is squared before multiplying by the coefficient.

When x = 3, I didn't do two lots of three and get six and square that to get 36, I did 3² and then multiplied that by two.

Be careful with your priority of operations when using equations like this one.

Pop those values into the table.

Do the same for the -x values, two lots of -1² will be 2, two lots of -2² will be 8, two lots of -3² will be 18, and we've got our populated table of values.

When we graph that, can you see what's changed? This is still a parabola, but a steeper one.

It's easier to see how steep it is when we compare it to the graph of y = x².

The parabola of y = 2x² is steeper.

Quick check you've got that.

Alex is plotting the quadratic equation y = 5x² - 2 and has made an error in the calculations of his coordinate pairs.

There's all of Alex's working out, and there's error amongst it.

Explain Alex's error to him before he goes on to calculate with the -x values.

Pause, have a look, what advice are we going to give Alex? Welcome back.

I wonder what you said.

You might have said, priority of operations means you square the x value before you multiply by five, not the other way around.

You see Alex has taken the x value multiplied by five, and then squared.

No Alex, priority of operations.

When we do it correctly, five lots of 0² - 2, that's -2, five lots of 1² - 2 is 3, and so on.

And we'll get those values in our table.

A negative coefficient will change the appearance of the graph.

Remember the linear graph y = 3 + x? It looks like that.

Contrast that with the linear graph y = 3 - x.

Changing the x coefficient from positive to negative means that new graph has a negative gradient.

So if that's what happens for linear graphs, I wonder what's going to happen for quadratic graphs.

Let's compare these two quadratics, y = 3 + x² and y = 3 - x².

y = 3 + x² will have that table of values.

y = 3 - x², we're gonna calculate that table of values.

When x = 0, 3 - 0² that's 3.

When we increase x to 1, 3 - 1² is 2.

When we increase x to 2, 3 - 2² is -1.

When we increase x to 3, 3 - 3² = -6.

Pop those in the table, something's different.

As we increased x previously, we saw the y value increase at the same time.

All of a sudden we've gone from x = 0, to x = 1, to x = 2, to x = 3.

And we're seeing the respective y values decrease.

That's interesting.

When we work out the y values for the respective -x values, our table looks like that.

Now let's contrast what the graphs look like.

There's y = 3 + x², a parabola, a y intercept as 0, 3, that's what we were expecting.

So what's gonna happen for y = 3 - x²? Let's look at those coordinate pairs.

Aha, something's different.

When we draw the parabola through there, that is the line y = 3 - x².

In the case of 3 + x², there's a positive x² coefficient.

Our parabola had a minimum y value, and then y values increasing in both x directions.

By contrast, y = 3 - x² is a -x² coefficient.

Our parabola has a maximum y value, and then y values decrease in both x directions.

When you see a -x² coefficient, your quadratic graph is going to look like that.

You might wanna pause here, maybe take a screenshot, maybe sketch those, and take a few notes, 'cause this is important.

Let's check you've got this.

If you've got it, you're gonna be able to match the equation to the correct graph.

Four equations, four graphs, pause, match them up.

Welcome back.

Let's see how we did.

Hopefully you matched B to our first graph.

That's x² + 10, D to our second graph, that's y = 10x², look how steep that parabola is.

A, y = 10 - x² was the third graph.

And C, y = x² - 10 was the final graph.

Oh look, y = 10 - x².

It's got a y intercept of 0, 10, and a -x² coefficient, which results in a parabola with a maximum y value and then y values decreasing in both x directions from there.

Practise time now.

Question one, I'd like to complete the table of values, and then draw the graphs of these quadratic equations.

There are three graphs for you to draw on that axes.

Pause and do this now.

Question two, Alex is plotting this quadratic equation and knows he's made a mistake with one coordinate pair.

He's plotting y = x² - 1, there's the table of values and that's what he's plotted on the graph.

For part A, I'd like you to write a sentence explaining how you immediately know which coordinate pair is wrong.

For part B, I'd like you to correctly re-plot the incorrect point, and then draw the graph of y = x² - 1.

Pause and do this now.

Question three, this table of values is correct for the quadratic equation y = 2x² - 10.

I'd like you to use any point in between these coordinate pairs to demonstrate why they should not have been joined with line segments like so.

Pause and do that now.

Feedback time now.

Let's see how we did.

y = x² - 7 was our first quadratic equation.

You should get those y values in your table, and then that graph when plotted.

It's y = x² - 7.

For the second equation, y = 5 - x².

You should have that table of values.

And that parabola.

For the last one, y = 10 - 2x².

There's your y values, and there's your graph.

You might wanna pause and check that your values and your tables, your coordinate pairs and your parabolas, are exactly the same as mine.

For question two, part A, write a sentence explaining how you immediately know which coordinate pair is wrong.

It's that coordinate pair.

How do we know? You might have said, this point doesn't fit the parabolic shape of the graph.

You might also have said, the symmetry in the y values is broken.

Look at the y values in that table, pair of eights, pair of zeros.

Surely we should also have a pair of threes.

You could have justified your answer like so.

Correctly re-plotting the incorrect points would've seen you plot the coordinate -2, positive 3.

The coordinate pairs and the parabola look like that.

For question three, I said to use any point in between the coordinate pairs to demonstrate why they should not have been joined with straight line segments.

I'll just take a few examples.

We could have chosen when x = 0.

5.

If we read from the graph there, we get the coordinate 0.

5, -9.

When we substitute that into the equation, the equation is not satisfied.

Could have picked any other point.

1.

5, -5 was on this particular line segment.

If I substitute the x value 1.

5 and the y value -5 into the equation, the equation is not satisfied.

That point should not be on this line.

How about a -x value, -2.

5, positive 3, I can read from this graph, substitute in those values.

Again, the equation is not satisfied.

Any point correctly read should have demonstrated this.

Onto the second part of the lesson now, where we're going to be plotting more complex quadratics.

Not all quadratics have the y axis as their line of symmetry.

Sounds exciting.

Let's have a look.

y = x² - 8x + 12.

Ooh, that equation is a little bit different.

When substituting a value into multiple terms in an expression, take extra care to check your accuracy.

When I substitute in x = 1, notice I'm gonna calculate 1², and I'm gonna calculate eight lots of one and I'm gonna add my 12 constants at the end.

I'm gonna do it step by step so I know when I do 1 - 8 + 12, I've absolutely got the correct y value.

I'll do the same for two.

2² - eight lots of 2 + 12, well that'll become 4 - 16 + 12.

That's going to be zero.

When x = 3, 3² - eight lots of 3 + 12, that's 9 - 24 + 12, step by step, careful as we go, and we get an accurate y value of -3.

Pop those into my table.

If you're permitted to use a calculator, then absolutely use it.

If I substitute in x = 4 into the equation on my calculator it looks like so, I get a y value of -4.

By putting the variable inside a bracket, it's easy to use the left arrow key to go back in and change it for the next value.

I'll use my left arrow key, go back, delete the four and change it to a five, and there we go, I've substituted in x = 5.

I'll get a y value of -3, do the same for x = 6, and do the same for x = 7.

There's our table of values.

When we plot the coordinate pairs and then join a nice smooth curve, still a parabola, but the line of symmetry is now the line x = 4.

How interesting.

Quick check you can do that.

I'd like you to populate this table of values and identify where the line of symmetry is for this parabola, y = x² - 5x + 4.

I'd like you to try calculating some values with a calculator, and some without.

There'll be moments where we're allowed to use a calculator, there'll be moments where we're not.

It's important to know that you can do both of those accurately.

Pause, complete the table of values.

Welcome back.

Hopefully you got those y values.

When we plot that quadratic, can you see where the line of symmetry is? A line of symmetry x = 2.

5.

How lovely.

Sofia is plotting this quadratic, y = x² + 5x + 4.

And she gets those values.

There's no symmetry in my table of values.

Maybe this quadratic does not form a parabola.

What do you think? Has Sofia found an exception? I don't see any symmetry in that table.

Perhaps she's right.

Pause, have a think.

See you in a moment.

Welcome back.

I wonder what you thought.

It is a quadratic equation.

It will form a parabola.

We just can't see it clearly in this region, we need more values, more coordinate pairs.

If we extend that table to include values all the way down to x = -6, and we look, starting to see a symmetry in the y values.

A pair of negative twos, a pair of zeros, a pair of fours, a pair of tens.

From there you can see this is gonna have a line of symmetry.

Sofia says, "It is a parabola, I know to check a wider range of values in the future.

Thank you".

You're welcome Sofia.

Quick check you've got this.

True or false? All quadratic equations will form a parabola.

Is that true or is it false? Once you've decided, I'd like you to choose one of the two statements at the bottom to justify your answer.

Pause and do that now.

Welcome back.

I hope you said true and used the statement B, all quadratic equations will form a parabola, you may need to look at more coordinate pairs to see this.

In the graph we have there, we haven't got enough coordinate pairs, we haven't got the correct range of values to see the symmetry, to see the shape of the parabola.

If we extended the graph, there we can see a beautiful parabola.

Sofia's having a problem with this equation, y = 2x² + 3x - 1.

I can see why Sofia's having a problem.

That's a tricky one.

Sofia says, "I see decreasing y values reaching a minimum, then increasing again".

Of course the y values go 8, 1, -2, -1, 4, 13, 26.

So Sofia says, "It could form a parabola, but either, one, that doesn't have a line of symmetry or I have made a mistake in some of my values".

What do you think? Has Sofia made errors? Will this curve not have symmetry? Pause, have a think.

Welcome back.

Let's see what's going on here.

You should always check your work in maths.

So we're gonna check Sofia's calculations.

Are those y values correct? If I check the first few on my calculator, I substitute in x = -3, we get positive 8.

I was expecting that.

When x = -2, y is positive 1.

That one's right.

When x = -1, y is -2, when x = 0, y is -1.

Sofia, you're right so far.

When x = 1, y = 4.

Oh no, my calculator ran out of batteries.

We'll do the rest manually.

When x = 2, two lots of 2² + three lots of 2 - 1, that would be 8 + 6 - 1, that is indeed 13.

What about when x = 3? 18 + 9 - 1, that's 26.

Sofia's values are absolutely right.

I don't see any symmetry in the y values.

However, when you plot those coordinate pairs, we get this parabola.

The values are all correct, all parabolas have symmetry, but in this case it's not a very obvious symmetry to see in either the table of values or on the graph.

The line x = -0.

75 is our line of symmetry now.

For many quadratics, the symmetry will not be visible in an integer table of values.

Sofia says, "This is cool.

There is so much more to quadratics than I thought".

You're right Sofia, there is.

Quick check you can do this accurately now.

I'd like you to complete a table of values for this quadratic, y = 4x² - 10x + 1.

Pause, work out those missing y values now.

Welcome back.

Let's see how we did.

I used my calculator to find the respective y value for when x = -2, I get 37, when x = 1, the y value is gonna be -5.

Let's do a few without the calculator.

When x = 3, my working out looks like that, when I do it carefully, step by step 36 - 30 + 1 gives me 7, when x = 4, my working out looks like that, carefully, step by step, 64 - 40 + 1 gives us 25.

Our table of values should look like so.

Let's put those coordinate pairs onto a graph and draw a nice smooth parabola.

And we see it's a parabola with symmetry in the line x = 1.

25.

Practise time now.

Question one, I'd like you to complete the table of values and plot the graph of this quadratic equation, y = x² - 3x - 4.

Pause and do this now.

For question two, I'd like to complete the table of values, plot the graph of this quadratic equation, y = -(x + 2)².

Pause, work out those values, plot the coordinate pairs, and draw a lovely parabola.

For question three, I'd like you to complete the missing values in this table and draw the graph of this quadratic, y = 5x² - 7x + 2.

Pause and do that one now.

Let's find out how we got on.

For question one, the quadratic y = x² - 3x -4, we should have that table of values, those coordinate pairs, and that parabola.

Maybe pause and just check your values match mine, your coordinate pairs match mine, and your parabola looks just like mine.

For question two, our table of values should look like so, and your parabola should look like that.

Again, pause, just check your work matches mine.

And for question three, those are the missing values in the table, there's the coordinate pairs plotted, and the parabola drawn.

Again, pause and check that yours matches mine.

Sadly, we're at the end of the lesson now, but we've learned that we can generate coordinate pairs for a quadratic graph from its equation, and then draw the graph.

The graphs of all quadratics, no matter how complex the equation, will form a parabola.

Hope you've enjoyed this lesson as much as I have, and I look forward to seeing you again soon for more maths.

Goodbye for now.