This maths hurts my braincells (X-intercept and Vertex)

celeste • 10 December 2012 at 9:59 PM

Ok, so, I have a question about finding the vertex of a parabola based on the x-intercepts.

The quadratic equation for this parabola is y=(x-1)(x+3), making the x-intercepts -1 and 3, right? (correct me if I'm wrong!)

The median of -1 and 3 is 1, and, to find the vertex, 1 must be substituted into the original equation (x-1)(x+3)

Making it (1-1)(1+3)

However, this is wrong, as the answer is 0.

Is there something that I'm doing wrong? I'm having a similar problem with y= x^2 - 4x -5 (which simplifies to (x-5)(x+1)

The answer is seven, but the given graph only goes up to positive 5 (and negative 10)

Help appreciated!

-Celeste

EDIT: Solved! But I have another question now...

"If the straight lines kx = 4y + 5 and (2k + 2)x = 7- 6y are parallel, find k"

savy839 • 12 December 2012 at 12:55 PM

@celeste
The x intercepts are what make the equation equal to 0
So you have -1 as an x intercept, but -1-1 is -2 and not 0 which is what you want it to equal. (that probably didn't make too much sense the way I worded it) same with 3. You said it was 3, but 3+3=6 and we're trying to get it to equal 0. So it's 3+(-3) and 1-1. In short, just switch the signs of the x intercepts you had.
Then you can take the median of 1 and -3 which is -1 and you get
(-1-1)(-1+3)
(-2)(2)
-4 so the vertex is -1, -4 😊

For the second one, I assume you had the intercepts as -5 and 1, but change those to 5 and -1 and you'll take the median which is 2 so subsitute 2 in to get
(2-5)(2+1)
(-3)(3)
(-9)
Hope that helped a bit, let me know if you need any more ^^

celeste • 12 December 2012 at 7:42 PM

@savy839
Thanks! 😊

Actually, I got it as soon as I posted that 😋 (I just looked through my notes)

However, I've got another question now ._.

"If the straight lines kx = 4y + 5 and (2k + 2)x = 7- 6y are parallel, find k"

(Sorry if the English is terrible; I've gotten used to it but you might have trouble understanding it)