1. ## finding x-intercepts: EASY

Sketch the graphs the following function:

f(x) = 10 + 3x -x^2

I know how you would go about sketching this. You would first have to find the x and y-intercepts however, I am having trouble finding the x-intercepts.

The thing is I've never factorised a quadratic where the leading, ax^2 is negative. Note: It's just the '-' in the x^2 that's confusing me. I've tried a few times with failure:

-x^2 + 3x + 10 = 0
-(x + 5)(x - 2) = 0

however, I think it's supposed to be factorised to (5 -x)(2 + x)
anyway, how would I go about contiuning to solve for the values of x for which the y values would equal 0? please outline each step slowly for me so I'll understand.

Thank you.

2. Hiya,

Both your initial answer, and the one you expected, were correct! They're the same thing written differently. But there is a much easier method.

If you start at

$-x^2 + 3x + 10 = 0$

Now add $x^2$ to both sides:

$3x+10=x^2$

Now take away $3x$ from both sides:

$10=x^2-3x$

Now take away $10$ from both sides:

$0=x^2-3x-10$

So if you just reverse all of the +/- signs, you'll get the same result for the intercepts. Isn't that much easier to factorize?

3. Originally Posted by Quacky
So if you just reverse all of the +/- signs, you'll get the same result for the intercepts. Isn't that much easier to factorize?
WOW, yeah thank you so much!
...it was so simple.

4. Originally Posted by Quacky

$-x^2 + 3x + 10 = 0$

So if you just reverse all of the +/- signs, you'll get the same result for the intercepts. Isn't that much easier to factorize?
Easier again, just multiply both sides through by $-1$

$-1 \times (-x^2 + 3x + 10) = -1\times (0)$

$x^2 - 3x - 10 = 0$