- #1

- 333

- 0

e.g.

0 = -x^2 + 4x + 21

0 = x^2 - 4x - 21

or

0 = -(x^2 - 4x - 21)

2) When you do simplify a quadratic by dividing it by whatever the a coefficient is, why do the x-values for the parabola remain the same the but the y doesn't? Does this mean you should only simplify quadratic equations to find x but leave it as is to find y-coordinates? For example for this equation I must find the x and y coordinates for the turning point, so I'm instructed to convert it into TP-form

y = 2x^2 - 12x + 22

y = x^2 - 6x + 11

y = x^2 - 6x + 9 -9 + 11

y = (x-3)^2 + 2

The y-coordinate is only half of what it should be because I divided by 2. Other than multiplying it by whatever I divided the equation by in the first place, is there another way to compensate for it? All my textbook says is to divide the equation but 'such problems are beyond the scope of this course'.

3) Last question :p

Am I factoring this correctly?

x^2 - 14x - mx + 1

x^2 - (14 + m)x + 1

If I were to plug 'b' into b^2, would it be -(14+m)^2?

-(14+m)^2

-(196 + 28m + m^2)

-196 - 28m - m^2

or do I have to times the negative one by everything inside first?

(-14-m)^2

196 + 28m + m^2

Would that mean that factoring the above expression like this:

x^2 +(-14 - m)x + 1 be valid?

Thanks