Video: Sketching Curves from their Roots

Solve π‘₯Β² βˆ’ 4π‘₯ + 4 = 0 by factoring, and hence determine which of the following figures would be a sketch of 𝑦 = π‘₯Β² βˆ’ 4π‘₯ + 4.


Video Transcript

Solve π‘₯ squared minus four π‘₯ plus four equals zero by factoring, and hence determine which of the following figures would be a sketch of 𝑦 equals π‘₯ squared minus four π‘₯ plus four.

We have five figures, one of which is a sketch of 𝑦 equals π‘₯ squared minus four π‘₯ plus four. The question is which one. We’re told that solving π‘₯ squared minus four π‘₯ plus four equals zero will help us. We write down the equation. Now how do we solve it? The answer is by factoring.

We want to write the left-hand side of our equation as a product of two factors, π‘₯ plus something times π‘₯ plus something. Let’s call those somethings that we need to find π‘Ž and 𝑏. To find the values of π‘Ž and 𝑏, we use the fact that π‘Ž times 𝑏 must be equal to the constant term of the trinomial, four.

This is a fact that is well worth remembering. I’ll explain why it’s true at the end of the video. If we assume that π‘Ž and 𝑏 are going to be integers, then we can just list all the possible values of π‘Ž and 𝑏 for which π‘Ž times 𝑏 is four. For example, π‘Ž could be one, in which case 𝑏 is four. Or π‘Ž and 𝑏 could both be two. It turns out that we don’t need to include the case that π‘Ž is four and 𝑏 is one.

This is just the first possibility with the values of π‘Ž and 𝑏 swapped. What we do need to consider is the possibility that π‘Ž and 𝑏 could be negative. This gives the other two possibilities, that π‘Ž is equal to negative one and 𝑏 is equal to negative four and that π‘Ž and 𝑏 are both equal to negative two. We’ve listed all the factor pairs of four including negative possibilities.

What do we do now? The other fact that we’ll prove at the end of the video is that π‘Ž plus 𝑏 is equal to the coefficient of the π‘₯ term, in our case negative four. For each column in our table, we calculate π‘Ž plus 𝑏. We thought that π‘Ž could be one and 𝑏 could be four. But this gives a value of five for π‘Ž plus 𝑏 and not negative four as we want. And so these aren’t the values that we’re looking for.

Similarly, when π‘Ž and 𝑏 are both two, then π‘Ž plus 𝑏 is four, not what we’re looking for. The next possibility gives negative five. It turns out that it’s the last option when both π‘Ž and 𝑏 are negative two. That gives us what we want. π‘Ž plus 𝑏 is then negative four. We found that π‘Ž is negative two and 𝑏 is negative two. And we can get rid of our working.

We need to substitute the values of π‘Ž and 𝑏 that we found. We get π‘₯ plus negative two times π‘₯ plus negative two equals zero. We can make things easier for ourselves by subtracting two instead of adding negative two. Okay, so our left- hand side is factored as required. You can check that distributing really does give π‘₯ squared minus four π‘₯ plus four.

These two expressions really are equal. Of course you might have just recognized that this is the perfect square π‘₯ minus two squared. We have factored as required. How does this help us solve our equation? We use the zero product property, which says that if the product of two factors is equal to zero, then one of those factors is equal to zero.

As both of our factors are π‘₯ minus two, we have that π‘₯ minus two must be equal to zero. And adding two to both sides, we see that our only solution is π‘₯ equals two. We have solved this equation. But how does this help us with our figures? We’re looking for the sketch of 𝑦 equals π‘₯ squared minus four π‘₯ plus four. And we’ve just solved π‘₯ squared minus four π‘₯ plus four equals zero.

Comparing these two equations, we can see that we’ve basically solved 𝑦 equals zero. Our answer was π‘₯ equals two and so π‘₯ equals two is when 𝑦 equals zero. On a graph or sketch, 𝑦 equals zero is the equation of the π‘₯-axis. And so we’re expecting an π‘₯-intercept at π‘₯ equals two. We see that option A has this π‘₯-intercept marked. When π‘₯ is two, 𝑦 is equal to zero.

However, we also see an π‘₯-intercept at π‘₯ equals negative two. However, we found that our equation only had one solution, π‘₯ equals two. π‘₯ equals negative two was not a solution of our equation. And so π‘₯ equals negative two should not be an π‘₯-intercept of our graph. For that reason, we can eliminate option A as a possibility. Option B however only has an π‘₯-intercept at π‘₯ equals two, as required.

This remains a possibility, as does option C. Option D has an π‘₯ intercept at π‘₯ equals negative two, not what we want. And option E, much like option A, has two π‘₯-intercepts and so isn’t the right sketch. Solving the equation allowed us to narrow down the options to a choice between options B and C. But we’re still going to have to decide.

One way of doing this is to consider the 𝑦-intercept, that is the value of 𝑦 for which the graph crosses the 𝑦-axis. In other words, the value of 𝑦 when π‘₯ is zero. We can substitute π‘₯ equals zero into the equation of our graph. We get that 𝑦 is equal to zero squared minus four times zero plus four. Zero squared and four times zero are both zero. And so we’re left with just the constant term four.

Looking at our two options, we can see that while option B has a 𝑦-intercept of four as required, option C has a 𝑦-intercept of negative four. We therefore eliminate option C and choose option B. Another way of choosing between options B and C would be to consider the coefficient of the π‘₯ squared term. This coefficient is one. One is greater than zero. And so we’re looking for an upward facing curve like we see in option B and not a downward facing curve like in option C.

To end this video, we’ll just prove the fact that we used earlier which allowed us to say that the product of A and B should be four and their sum should be negative four. We can distribute these two brackets. We can distribute the terms in these parentheses. We get π‘₯ squared plus π‘₯ times 𝑏, which is 𝑏π‘₯, plus π‘Ž times π‘₯ plus π‘Ž times 𝑏. We can swap and combine these like terms and we get π‘₯ squared plus π‘Ž plus 𝑏 times π‘₯ plus π‘Žπ‘.

We want to find values of π‘Ž and 𝑏 for which these two expressions are equal for all values of π‘₯. For these two expressions to be equal, their constant terms must be equal and so π‘Žπ‘ must equal four. This was the fact that we used to list several possibilities for the values of π‘Ž and 𝑏 in the integers.

And also the coefficient of π‘₯ of these two expressions must be equal. And so π‘Ž plus 𝑏 must equal negative four. This is what we used to choose between the possibilities that we had in our table. Having then found the values of π‘Ž and 𝑏, we wrote the equation of the graph in this form.

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