Math - Intercept Form

In this lecture we were introduced to the intercept form, or factored form, of the quadratic equation.

This form is incredibly useful for quickly seeing where the curve crosses the x-axis - hence it’s name.

For your challenge, find the equation in intercept form for the curve that runs through the points (2,0), (5,3), and (8,0).

Remember that this will be in the form; y = a(x - p)(x - q)

‘p’ and ‘q’ should be easy to plug in. You can then find ‘a’ by solving the equation using the other point.

Post your answers below and remember to use the spoiler tags.

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For a the points (1,0), (5,0) and a vertex of 3.
3 = a(3-1)(3-5)
3 = a * 2 *-2
3 = -4a
a = -(3/4)

For a the points (2,0), (8,0) and a vertex of 3.
0 = a (x-2)(x-8)
so p = 2 and q = 8
3 = a (5-2)(5-8)
3 = a * 3 * -3
3 = -9a
a = -(3/9)
a = -(1/3)

One thing I don’t get is that we need x-p and x-q to both equal 0. I mean, since it is 0 = a(x-p)(x-q), if
“a * 0 * (x-q)” for example, q can be anything and it will be equal to 0. The same for p. Isn’t it?

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@ProvencalG, hopefully this will become clearer as you progress through the rest of the section but here’s a brief explanation.

p and q help represent the x-intercept points, so (x - p) and/or (x - q) need to equal 0 when y = 0, so that the equation can balance.

If you have a parabola with two roots then there are two valid solutions for x when y = 0 but you can only plug one of your answers into the equation at a time, so you can kind of think of this form as a way to wrap up two equations in to one.

0 = x - p
0 = x - q

Depending on which value you chose for x, either the first or second term in y = (x - p)(x - q) will become 0 and then the multiplication cancels out the other, allowing you to condense the two equations into one.

I hope that helps clear things up for you but if you’d like me to explain further, or feel that the lecture should be amended, please do let me know.


I think this one went better:

  1. 0 = -3/4 (x-1)(x-5)

  2. 0 = -1/3 (x-2)(x-8)

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I got:

y = -4/3(x-1)(x-5)


y = -3(x-2)(x-8)

Also p and q are the x values of the roots, is that right? I can’t recall whether that was explicitly stated in the video.

@Ajai_Raj, you got the parts inside the parentheses correct but not the coefficient ‘a’ on the outside.
You are incredibly close though and it looks like you made the same small error in both cases.

Here are the correct answers to check your workings…

The first one is:
3 = a (3 - 1) (3 - 5)
3 = a * 2 * (-2)
3 = -4a
a = -3/4
Therefore, y = -3/4 (x - 1) (x - 5)

And the second one is:
3 = a (5 - 2) (5 - 8)
3 = a * 3 * -3
3 = -9a
a = -3/9
a = -1/3
Therefore, y = -1/3 (x - 2) (x - 8)

In answer to your question about p and q, I cover these in more detail in the lecture “Factoring Quadratics”.
They’re not really the x-value of the roots themselves. Instead they’re the amount you need to add to the x-value of the roots in order to make the parentheses equal zero.
So if you had roots at (-2,0) and (3,0) p and q would be be 2 and -3 respectively.

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Ah, went back and got it the second time around. Thanks @garypettie!

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Very shorthand writing…
My maths professor wouldn’t approve of this… :speak_no_evil:

y = a(x-p)(x-q)
p=1, q=5
3 = a(3-1)(3-5)
3 = a2-2

p=2, q=8

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This ‘p’ and ‘q’, intercept form, I totally forgot over the years. :wink:

Fun stuff.


For the intercept at points (1, 0), (3, 3), and (5, 0):


For the version through (2,0), (5,3), and (8,0):



I’m not making the leap completely yet - I understand that one of the x values needs to equal zero, and I read the response about wrapping both up into one equation - I’m hoping that the next lecture or two will cause it to click completely (It feels close, like I’m almost getting the why). I never really remembered formulas when I took math until I started calculus and found out where the formulas came from. But it’s so far removed - my brain is reaching trying to remember the whys still :slight_smile:

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Great work @Dendrolis.
We cover a few different ways of writing quadratics in this section and each reveals something about the parabola we’re dealing with.
Hopefully things will all start to click into place for you over the next few lectures but if you still have questions, please don’t hesitate to ask.

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for the first one y = a(x-p)(x-q)
0 = a(1-1)(5-5)
3 = a(3-1)(3-5)
3 = a(2)(-2)
3 = a*-4
a = 3/-4
a = -0.75 (or -3/4)

for the second oney = a(x-p)(x-q)
0 = a(2-2)(8-8)
3 = a(5-2)(5-8)
3 = a(3)(-3)
3 = a * -9
a = 3/-9
a = 0.33333333 etc or (-1/3)

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For the first challenge I got

0 = -3/4(x-1)(x-5)

For the second challenge I got

0 = -1/3(x-2)(x-8)

But, I didn’t understand why we divide by 2 to get the X component of the vertex.

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@Shahari93, a parabola is perfectly symmetrical, so you can find the x-component of the vertex by adding the two known x-components of the roots and then halving the result.

For example, if you had roots at (4,0) and (6,0), we would expect the highest point of the parabola to exist at (5, ?) due to this symmetry.

If we do the math, we get:
(4 + 6) / 2 = 10 / 2 = 5

So both our assumption and the math are in agreement.

I hope this helps clarify things for you but if you have any other questions please ask.

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