Writing a quadratic function from solutions | Algebra 1 (TX TEKS) | Khan Academy

We're told a quadratic function ( f ) has two real solutions ( x = -3 ) and ( x = 5 ) that make ( f(x) = 0 ). Select the equations that could define ( f ) in standard form. So, pause this video and have a go at that before we do this together.

All right, so there's a bunch of ways you could approach this, but the way that I think about it is we can express this quadratic in terms of its two solutions. So, you could have ( x - ) the first solution, and the first solution here is when ( x ) is equal to -3, and then times ( x ) us the second solution when ( x ) is equal to 5.

Now, why does this work? Well, think about it. If ( x ) is equal to -3 right here, and if I were to subtract another -3, well then this is going to be equal to 0. ( 0 \times ) anything is zero, and then ( f(-3) ) would be zero. Similarly, if ( x ) were equal to five here, well then this whole thing would be equal to zero; ( 0 \times ) anything is 0, so ( f(5) ) is zero.

Now, this is a definition of the quadratic, but it is not in standard form. Standard form, as a reminder, would be some constant times ( x^2 ) plus some other constant times ( x ) plus some other constant. So, to get there, we have to multiply this out.

And actually, before we do that, let me just simplify a little bit. This is going to be equal to ( x ) when I subtract a -3. That's the same thing as adding three, and then times ( x - 5 ). So, ( x + 3 \times x - 5 ).

And now we can expand this out so we get it to standard form. So, this is going to be equal to ( x \times x ), which is ( x^2 ). We have ( x \times -5 ), which is -5x. We have 3 times ( x ), which is 3x, and then we have 3 times -5, which is -15.

So, last but not least, we have ( x^2 ), and if I am subtracting 5x and then I add 3x, that is -2x minus 15. So, this is ( f(x) ) in standard form.

Now, let's see which of these choices gets me this. So when I look over here, well, what's interesting is all of these have a coefficient of either 2 or -2. I don't see that over here. So what is happening here is I can multiply this whole thing by 2 or -2, and it's not going to change where my zeros are.

Why is that? Well, think about it. If I had a 2 over here, when ( x ) is equal to 5, this is going to be ( 0 \times ) something (\times 2); it's still going to be equal to zero. Similarly, if that were a negative -2, so I'm going to have the same zeros if I multiply it by really any number that is not zero.

So let's do that. If I were to multiply this equation by positive 2, I need to multiply all of them by two. I'm running out of space, so I'll do it up here. We would get ( f(x) ) is equal to ( 2 \times x^2 ), which is ( 2x^2 ); ( 2 \times -2x ) is -4x; ( 2 \times -15 ) is -30. That's one way we could think about it.

Another way we could say maybe ( f(x) ) is going to be equal to, and to be clear, these are not the same functions. When I multiply it by 2 or -2, it does fundamentally change the function, but they would have the same zeros; they would have the same two real solutions ( x = -3 ) and ( x = 5 ).

So if I were to say, “Well, maybe instead of this, ( f(x) ) could be this times -2,” once again, it's a different ( f(x) ); it's a different function. In these situations, I'm just trying to find out all the possibilities, and there could be many more. I could multiply it by 3 or -3 or anything else.

But if I were to multiply this by -2, I would get ( -2x^2 ); ( -2 \times -2x ) is ( +4x ); ( -2 \times -15 ) is ( +30 ). So I’m going to say it one more time, the three things that I’m boxing off here, these three possible functions, these are all different functions. If I were to graph it, they would all look different, but they all have the same two real solutions ( x = -3 ) and ( x = 5 ).

So now, let's see which choices match up: ( 2x^2 - 4x - 30 ), ( 2x^2 - 4x - 30 ). I like this one right here and then ( -2x^2 + 4x + 30 ), ( -2x^2 + 4x + 30 ). I like this one here as well, so I'm done.