Lesson video

Hello, Mr. Robson here.

Welcome to maths.

Superb choice to join me.

Today we're looking at the key features of quadratic graphs and quadratic graphs are awesome.

So let's get started.

Our learning outcome is that we'll be able to identify the key features of a quadratic graph.

Lots of keywords in today's lesson.

Parabola.

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

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

Quadratic, that's another keyword.

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

The general form for quadratic is axe squared plus bx plus c.

Roots is another keyword.

When drawing the graph of an equation, the roots of the equation is where its graph intercepts the x-axis where y equals zero.

For example, this quadratic has roots at x equals two and x equals five.

Turning point.

The turning point of a graph is a point on the curve where as x increases, the y values change from decreasing to increasing or vice versa.

Two parts of today's lesson and we're gonna begin with the key features.

Linear graphs in the form y equals mx plus c can be described uniquely by their key features.

For example, this linear equation, y equals 3x minus four, it's a straight line graph.

It's got a gradient of positive three, a y-intercept at zero, negative four and an x-intercept at negative four over three, zero.

Quadratic graphs also have key features which uniquely describe them.

We can see the key features of a quadratic graph in this example, Y equals x squared minus 2x minus three.

It forms a parabola.

It has a y-intercept at zero, negative three.

It has 2x-intercepts at negative one, zero and three, zero and it has a turning point at one, negative four.

Let's take a look at some of those key features, Y-intercept at zero, negative three.

A graph intercepts the y-axis when x equals zero.

For this graph, that's there, x equals zero, y equals negative three.

We have 2x-intercepts at coordinates negative one, zero and three, zero.

A graph intercepts the x-axis when y equals zero.

That's here in this case.

We call the x values the roots of the equation.

So for this quadratic equation, we'd say it has roots at x equals negative one and x equals positive three.

We also have a turning point at one, negative four.

The turning point of this quadratic is where the graph goes from having decreasing y values to increasing y values.

The parabola turns around if you will.

We call that moment a turning point.

In this case, one, negative four.

There's something else important about this turning point.

We call this turning point the minimum point.

That's because there will be no lower y value than negative four on this graph.

Quick check you've got that.

I'd like to fill in the blanks for the key features of this quadratic.

Pause and do that now.

Welcome back.

I hope you filled it in like so.

The graph of a quadratic equation forms a parabola.

That's how we describe the shape.

It's a keyword, parabola.

This quadratic has roots at x equals negative four and x equals two.

Those roots are where y equals zero where we intercept the x-axis.

We say it's got roots at x equals negative four and x equals two.

What was special about that coordinate pen negative one, negative nine? We have the turning point.

Hope you used those keywords there.

That's the turning point.

And then finally, that means that negative nine is the minimum y value.

So I'm not saying is that a turning point, it's a minimum point on the graph.

Andeep notices a key difference when plotting this quadratic.

Y equals four minus x squared.

Andeep says, "It forms a parabola.

I see roots, I see a turning point, but there's something different about this quadratic.

Can you spot it?" Pause and have a think.

What's different? Welcome back.

I hope you said something along the lines of what Andeep said.

"This turning point is a maximum point.

We go from having increasing y values to decreasing y values.

In this case, four is the maximum y value." It's key that we understand this.

The quadratic on the left, y equals x squared minus 2x minus three is a quadratic with a positive x squared coefficient, so it forms a parabola with a turning point, which is a minimum.

Contrast that with y equals four minus x squared.

It's a quadratic with a negative x squared coefficient.

Therefore, we have a parabola with a turning point, which is a maximum.

It's important you know the difference between these.

You might wanna pause and make a few notes.

Let's check you've got this.

I'd like you to fill in the blanks with the given words and equations.

Pause and see if you can allocate those labels now.

Welcome back.

Let's start by labelling the roots of each graph and then needs to label the turning points.

It's important we differentiate between a turning point which is a minimum and a turning point which is a maximum.

So which quadratic is which? On the right-hand side, we have the quadratic y equals one minus 6x minus x squared.

On the left-hand side is the quadratic y equals x squared minus 10x plus 15.

What's the difference? One has a positive x squared coefficient.

Therefore, the turning point is a minimum.

One has a negative x squared coefficient.

Therefore, the turning point is a maximum.

Andeep is exploring this quadratic now.

For part A, Andeep's asked to plot y equals x squared minus 6x plus eight.

For part B, Andeep is asked to solve x squared minus 6x plus eight equals zero.

Oh look, it's the same quadratic.

Thankfully, Andeep's pretty awesome at maths, so he quickly populates a table of values and draws the parabola.

And then when it comes to solving, Andeep spotted it factorises.

If it factorises like, so we've got two solutions, a equals two and x equals four.

And Andeep says, "I see a link between the key features of the graph and the solutions to the equation.

Can you see it too?" Pause.

Can you spot a link between the solution to the equation and the key feature of the graph? Welcome back.

I wonder if you noticed what Andeep noticed.

Andeep says, "The solutions to the equation give us the x coordinates of the points two, zero and four, zero.

This means if you can solve an equation, you have found its roots." The roots of the equation are found where the graph intercepts the x-axis, i.

e.

when y equals zero.

In this case, y equals zero when x equals two and x equals four.

We can see that both in the solution to the equation and on the graph of that parabola.

So in this case, we've got roots at x equals two and x equals four.

Quick check you've got that skill.

Which of these is the correct factorization of this quadratic? Y equals x squared minus 2x minus 15.

Three to choose from.

Pause and take your pick.

Welcome back.

I hope you went for option C.

That quadratic factorises to x plus three, x minus five.

Now that we've factorised it, my next question for you is which of these are roots of this quadratic? Now that we factorised, we can solve.

When does y equal zero? Therefore, what are the roots? Pause and take your pick from those three options.

Welcome back.

Let's consider each of those options.

You might have put A.

That's because a root is located here.

At the coordinate negative three, zero, we would see a root.

However, that's not how we communicate a root.

We would just say x equals negative three is one of the roots.

We wouldn't say the full coordinate pair.

B was not a solution to x plus three, x minus five equals zero.

Y will be equal to zero when x equals negative three, not positive three.

C was absolutely right.

Y will equal zero when x equals five, making this a root.

Sometimes quadratics don't factorise, in which case we can use the graph to estimate their roots.

Here's the graph of y equals 2x squared minus 9x plus eight and I could say the roots are there and then estimate the x value to one decimal place.

I would say the quadratic has estimated roots at x equals 1.

2 and x equals 3.

3.

Quick check you can estimate roots too.

I'd like you to estimate the roots of this quadratic to one decimal place.

Pause and do that now.

Welcome back.

I hope you spotted those two intercepts of the x-axis and estimated the roots to be x equals 0.

3 and x equals negative 0.

8.

Andeep is exploring another quadratic.

Y equals x squared minus 6x plus nine, and there's a table of values for the coordinate pairs.

Andeep says, "I found something really cool! This quadratic is different to the ones I've graphed before." Can you see what's different about this quadratic? Try and imagine where those coordinate pairs are going to be on that graph and see if you notice something different.

Pause.

Have a think about that now.

Welcome back.

I wonder what you spotted.

Did you notice this? When we plot those coordinate pairs and draw a parabola, this quadratic only has one root.

Because x squared minus 6x plus nine is a perfect square, it factorises to x minus three squared.

Therefore, there's only one value that makes y equal zero.

Therefore, there's only one root.

When we have a case like this where there's only one root, the parabola intercepts the x-axis just once.

We call it one repeated root.

You might wanna pause and write that down.

Andeep is exploring another quadratic.

It's a tiny change to that equation now by looking at the equation y equals x squared minus 6x plus 10 with that table of values.

Andeep says, "I found another quadratic which is different again!" Can you see what is different about this one? Pause.

Try and imagine where that parabola is going to be.

See if you can spot a difference.

Welcome back.

What did you notice? Hopefully you imagined the quadratic would look like that.

In which case this one has no roots.

There's no x value which makes y equals zero.

Therefore, the parabola will never intercept the y-axis.

This quadratic has no roots.

Quick check you've got that.

I'd like you to match the descriptions to the graphs.

Do that now.

Welcome back.

Hopefully you said the first graph represents a parabola with two roots, the second graph, no roots, and the third graph was an example of one repeated root.

Practise time now.

Question one, for this quadratic equation for part A, I'd like you to complete the table of values and draw the graph.

For part B, I'd like you to identify the roots.

For part C, I'd like to identify the turning point and write a sentence to justify why it is a minimum point.

Pause and do this now.

Question two.

This is the graph of y equals negative 2x squared plus 8x minus six.

Write a description of the graph.

Be sure to mention all of its key features.

You should write at least three sentences for this.

Pause and do that now.

For question three, for this quadratic equation, I'd like you to complete the table of values and draw the graph.

For part B, I'd like to use your graph to estimate the roots of the equation.

Pause and do this now.

Question four, I'd like you to factorise to identify the roots for each of these quadratics.

Pause and give these a go now.

Let's see how we got on.

Question one, completing the table of values for this quadratic equation should have looked like that, which will give you that graph.

Once you've got that graph, you can identify the roots.

The roots are x equals one and x equals seven, and a turning point is at four, negative nine.

You should have written something along the lines of the turning point at four, negative nine is a minimum.

Y values decrease to that point, then increase afterwards.

Negative nine is the minimum y value.

Question two, I ask you to write a description of the graph being sure to mention all of its key features.

You might have written the graph forms a parabola with roots at x equals one and x equals three.

The turning point at two, two is a maximum.

The curve has a y-intercept at zero, negative six.

Question three, completing the table of values would've given you those two missing y values.

When plotted, the graph looks like this, and then we can estimate the roots to be at x equals negative 2.

7 and x equals 0.

2.

For question four, I ask you to factorise to identify the roots.

Part A factorises to x plus four, x plus five, so we'll have roots at x equals negative four and x equals negative five.

B factorises like so, giving us roots at x equals seven and x equals negative two.

C factorises like so, giving us roots at x equals negative four and x equals positive four.

And D is a perfect square, factorising like so.

It's only got one repeated root at x equals negative five.

Be sure to write one repeated root next to your x equals negative five.

Onto the second part of the lesson now where we're going to be locating the turning point algebraically.

What is the turning point of y equals x squared? That's quite a simple question.

It's zero, zero.

More importantly, why is zero, zero the turning point? It's because x equals zero gives us the lowest possible value of the x squared term and that value is zero, hence zero, zero.

When x equals negative one or x equals positive one, we have x squared equals one.

When x equals negative two or positive two, x squared equals four.

Every other x value than zero gives us a higher y value.

That is why the turning point is at zero, zero.

What if we change the equation slightly? Y equals x squared minus one, a turning point at zero, negative one.

Why is that the turning point? It's because in x squared minus one, the only term we can affect is that of x squared.

Once again, x equals zero will give us the lowest value for x squared.

If x squared equals zero is the minimum value of that term, then y equals negative one is the lowest possible y value, hence the turning point is at zero, negative one.

Let's change that equation ever so slightly again.

Let's turn it into y equals x minus three squared minus one.

What's the turning point now? It's at three, negative one.

Why is that? It's because in the expression x minus three squared minus one, the only term we can affect is that of x minus three squared.

Squaring will always give us a positive value unless we are squaring zero.

Therefore, the minimum value occurs when x minus three equals zero.

That is when x equals positive three.

When x equals positive three, y equals negative one.

That's why we get a turning point at three, negative one.

Quick check you've got that.

What is the turning point of y equals x plus four squared minus three? Three coordinate pairs to choose from.

Pause.

Have a think.

Welcome back.

I hope you went for C, negative four, negative three.

It's a minimum point.

There can be no lower y value than negative three.

Y is the turning point of y equals x plus four squared minus three.

The point negative four, negative three.

Which of these three justifies that minimum point? Pause and have a think about those three now.

Welcome back.

I hope you were courageous enough to say all three of them.

Why is that the turning point? It's because the only thing we can affect is x plus four squared and the minimum we can make it is zero.

X plus four squared is zero when x equals negative four.

That's why x equals negative four is the x coordinate of that minimum point.

And when x plus four squared is zero, y is equal to negative three, hence, x equals negative four, y equals negative three is our turning point and the minimum point on that graph.

The turning point of y equals x minus three squared minus one is the point three, negative one.

Andeep says, "I recognise this; it's completing the square.

Putting a quadratic in the form y equals x plus p squared plus q." If we look at our equation, expand that bracket and simplify, Andeep says, "This is the quadratic y equals x squared minus 6x plus eight in completing the square form." And he's correct.

What this means is if we can write a quadratic equation in the form x plus p squared plus q equals zero, we can identify the turning point.

If we look at y equals x squared minus 14x plus 40 and consider what we've got, we've got an x squared, negative 14xs and positive 40.

We haven't got a complete square, we haven't got x minus seven squared.

In fact, we're nine short.

So when I write x squared minus 14x plus 40 in completing the square form, I have x minus seven squared minus nine.

From here, we know the minimum value of x minus seven squared is zero when x equals positive seven, and if x minus seven squared is zero, then the minimum value of y is negative nine.

Hence, we have a turning point of seven, negative nine.

We can draw the graph of that just to be sure that we're correct.

There we go.

A turning point at seven, negative nine.

When we write a quadratic equation in that form, the turning point is negative p, q.

We know this because the minimum value of x plus p squared is zero and it occurs when x equals negative p.

When x plus p squared is zero, the value of y is q, so the turning point must be negative p, q when we have an equation written in that form.

Andeep has noticed something else very useful.

For this case we saw a moment ago, Andeep says, "I like this method even more because I can go on to find when y equals zero which will give me the roots." Once we have it in completing the square form, we can rearrange.

Take the positive and negative root of nine, add positive seven to both sides and we'll get solutions x equals four, x equals 10.

Oh look, that's where we intercept the x-axis.

We've got roots at x equals four, x equals 10.

That's why it's really useful to rearrange your quadratic into completing the square form.

We can find both the turning point and the roots.

Quick check you've got that.

Which of these expresses the quadratic y equals x squared minus 4x minus 12 in the form x plus p squared plus q? Pause and take your pick now.

Welcome back, I hope you said it's B, y equals x minus two squared minus 16.

Now that we have that quadratic expressed as x minus two squared minus 16, what is the turning point? Pause, take your pick from those three coordinate pairs.

Welcome back.

I hope you went for option A, two, negative 16.

We will come across slightly more complex cases like this one.

If we wanna write that in completing the square form, we need an x squared coefficient of one, so we factorise.

We take out a factor of negative one.

Then we can write it in the correct form.

From here, let's multiply out those brackets.

Negative one lots of x minus three squared and negative one lots of negative nine, giving us positive nine.

From here, we know we'll have a minimum value when x equals three because we want x minus three squared to be zero.

When that is zero, y will be equal to nine.

So we have a turning point at three, nine.

That's correct, but there's something different about the turning point in this case.

Can you see what that is? Pause and have a think.

Welcome back.

I wonder if you spotted it.

It might help to rearrange the equation.

If we write it as y equals nine minus x minus three squared, you'll notice if x minus three squared is not zero, it must be a positive value.

What this means is our y values are going to decrease.

If we test the x integer values either side of three, when x equals two, y will equal eight, when x equals four, y will equal eight.

The turning point is a maximum point in this case.

I can draw the graph of it to justify why that is so.

A turning point at three, nine being a maximum.

But you knew it was going to be a maximum when you first saw the equation.

Y equals 6x minus x squared.

It's a negative x squared coefficient so you knew that was going to be a maximum point.

Another slightly more complex case.

Y equals negative 2x squared plus x minus eight.

We'll write this one in the form a, bracket, x plus p squared plus q.

We need an x squared coefficient of one to factorise out negative two.

We then take the content of that bracket and write it in completing the square form, which will look like so, and then we're gonna multiply out that bracket, negative two lots of x minus 1/4 squared and negative two lots of 63/16.

Once we're in this form, we know our turning point.

It'll be at when x equals positive 1/4, which will make x minus 1/4 squared, zero.

When that's zero, the y value must be negative 63 over eight and we know that this turning point is a maximum because we had a x squared coefficient of negative two to begin with.

Now, I think I'm right at this point, but it's always good in maths to know you're right.

For a complex case like this, it's a smart idea to use tech to check, so I went to desmos.

com and I drew the graph.

Now, Desmos uses decimals rather than fractions, but it's confirmed 1/4, negative 63 over eight is indeed the turning point.

When you're working on examples like this, it's always a good idea to use tech to check.

Another thing we now know is by demonstrating that the turning point is below the x-axis and it's a maximum, we've also shown that this quadratic has no roots.

Well done, us.

Quick check you've got that now.

I'd like you to write y equals 5x squared minus 9x plus four in the form a, bracket, x plus p squared.

Hence, find the turning point and declare how many roots the quadratic has.

Pause.

Take a few minutes to work through this problem.

Welcome back.

Hopefully you took our equation and wrote it like five lots of x squared minus 1.

8x plus 0.

8.

We could have used fractions here.

I'm using decimals 'cause it happens to work quite nicely in this case.

Turn that into completing the square form.

Simplify, I'm gonna multiply out that bracket, five lots of x minus 0.

9 squared and five lots of negative 0.

01.

The turning point must be at 0.

9, negative 0.

05 and it's a minimum because five lots of x minus 0.

9 squared will always be zero or greater.

Once we know it's a minimum turning point and it's below the x-axis, it means this quadratic will have two roots.

Practise time now.

Question one, I'd like you to write each quadratic in completing the square form, thus locate their turning point.

Pause and give those three a go now.

Question two, for each quadratic, locate the turning point and then write a sentence to justify if it is a maximum or a minimum.

Write a sentence to justify how many roots the equation has.

Three equations, two sentences for each.

Pause and do that now.

Let's see how we did.

For question one, I asked you to write each quadratic in completing the square form, thus locate their turning point.

We should have rewritten x squared minus 14x plus 61 like so and located a turning point at seven, 12.

For B, we should have rewritten it like so.

It was a perfect square.

Therefore, we have a turning point at nine, zero.

For C, we should have rewritten it as two lots of x plus 12 squared plus 13, giving us a turning point at negative 12, 13.

For question two, part A, we should have rewritten the equation in that form and from there, located a turning point at negative 15, negative 65.

And your sentence to justify whether this is a maximum or a minimum could look like this.

X plus 15 squared is always greater than or equal to zero.

Therefore, y equals negative 65 is a minimum value.

From there, we can say the minimum point is below the x-axis.

Therefore, there are two roots.

For part B, we should have rewritten that equation like so.

Found a turning point at three, 10.

Then your sentence to justify that this is a maximum could look like so.

Negative x minus three squared will always be less than or equal to zero.

Therefore, y equals 10 is a maximum value.

Once we know there's a maximum point and it's above the x-axis, we know that there are two roots.

For part C, we should rewritten that equation like so to declare there's a turning point at five, zero.

From there, we can say four lots of x minus five squared will always be greater than or equal to zero.

Therefore, y equals zero is a minimum value.

If the minimum point is on the x-axis, we therefore have one repeated root.

We're at the end of the lesson now and we have learned a lot.

We've learned that we can identify the key features of a quadratic graph.

A quadratic curve is a parabola and the quadratic equation can have zero, one or two roots.

A key feature of a quadratic graph is its turning point.

This can be identified and we can justify whether it is a maximum or a minimum point on the graph.

I hope you found that maths super interesting and I look forward to seeing you again soon for more.

Goodbye for now.