# Thread: 2nd degree polynomial help needed (parabola)

1. ## 2nd degree polynomial help needed (parabola)

By factoring, I found that (-3, 4) is something (vertex?), but after that I'm lost and need some help.

2.) Let y = f (x) = -x^2 + 7x – 12.
(If any of the questions below cannot be answered, write ‘NONE’ in the blank provided)
a.) Give the ordered pair for the y-intercept: ________________________

b.) Give the ordered pair(s) for any x intercepts: ____________________

c.) List all x-values that give y a value of 2: ________________________

2. a)To find the y intercepts you need to factor it out to find the zeros. You're answer isn't quite right. (look at the power of your 3)

b)When you have a function of x f(x) think of it as your y value. So you can think of your coordinates as $(3.5, f(3.5))$ so your y value will be the answer when you plug 3.5 into the formula.
if you want to find the x-intercepts.. the point were x=0 just plug it in and go: $f(0)= -(0)^2+7(0) -12$

c) to find your highest point you can imagine the two (x values) at the y intercepts. Because one is going up and the other is going down, so the point between will be the highest. Plug it into your f(x) equation and see how high it will go. Hint: look at your answer to a) and my example numbers in b).

3. Actually, the post above is mostly incorrect.

For starters, the point(s) I suppose you were attempting to find by factoring were the x-intercepts or the points on the graph where the parabola crosses the x-axis (abscissa), in other words where $y=0$.

The vertex (h,k) is taken from an equation already in vertex form $y=a(x-h)^2+k$

or, it could be derived from a function in standard form $y=ax^2+bx+c$

by using the formula $h=\frac{-b}{2a}$

k is determined by substituting h for x in the equation and solving for y (k)

a) the y-intercept is when $x=0$

substituting 0 for x yields: $y=-0^2+7(0)-12$

b) the x-intercepts, as stated before, can be found by factoring

some students find it easier to factor out -1 before factoring the trinomial, which would give: $y=(-1)(x^2-7x+12)$

c) to find the x-values where $y=2$

set the expression $-x^2+7x-12$ equal to 2 and solve for x: $2=-x^2+7x-12$

hint: you can check your work by graphing