In the solution video, the x-intercepts for the equation y = x(x - 10) + p were said to be x = 0 and x = 10
But this doesn’t work for every quadratic equation.
Eg. y = x^2 + 2x - 3
=> y = x(x + 2) - 3
x-intercepts = 0 or -2 (which is wrong)
In which cases is it okay to find the x-intercepts like that?
If the question was y = x(x + 2) + p, the answer would not be -3.
There was a typo in my example. I’ve fixed it
Where?
Earlier it was “=> y = x(x + 2) - 8”
Now it is “=> y = x(x + 2) - 3”
In the video solution, for the equation y = x(x-10) + p , the x-intercepts were considered to be x = 0 and x = 10.
I don’t understand how it applies to only that quadratic equation but not all other quadratic equations.
Why do you think the presenter made that example? To be clear, he isn’t claiming that this is true for every quadratic.
The goal was to get the lowest point of the parabola to figure out how many units we need to move up vertically so that there are no x-intercepts, but if the x-intercepts initially are incorrect, how can we be sure we found the right lowest point?
I don’t think you’ve gotten the spirit behind the question.
The presenter started with the example
y = x(x - 10)
That quadratic is as shown in the diagram, with x-intercepts 0 and 10 with the minima (5, -25).
Now how does that help in the question? Given that
y = x(x - 10) + p
if I set p = 25, that’s an upward translation by 25 units, making the minima (-5, 0). Which helps a lot in the context of this question.
Going back to your question, start with y = x(x - 2) instead. y = x(x - 2) - 3 is irrelevant.
