unit 4 quiz part 2 Flashcards
writing a quadratic equation when given the roots
x2 - sum of rootsx + product of roots = 0
to find a double root,
need b2 - 4ac= 0, (discriminant) (one will likely have the x value, so keep it in and then solve for x)
standard form for quadratics
ax2 + bx + c
vertex form for quadratics
y= a(x-h)2 + k (vertex is h,k)
Dx is
ALWAYS all reals of (-infinity, infinity)
Ry
EITHER y greater than or equal to the min or less than or equal to the max of the parabola
AOS
x coordinate from vertex, can also be found from -b/2a, and in vertex form, the x = h
set equation to zero and solve for x
solving for zeros/x intercept
plug in x as zero and then solve for y
finding the y-intercept
-only use x-values from the graph
- open intervals with (,)
intervals/maxs and mins of graphs
find sum/product of roots if they’re not provided
sum: -b/a
product: c/a
**roots cannot have denominators so use LCM and multiply them out
b2-4ac >0 (NOT a perfect square)
Real, Irrational, Unequal
- graph crosses x acis twice in two irrational points
b2-4ac > 0 (perfect square)
real, rational, unequal
- graph touches x axis in two rational points
b2-4ac = 0
real, rational, equal (double root)
- graph touches the x axis in one point; tangent to x axis
b2-4ac <0
imaginary, the parabola does not touch the x axis
when asked to describe nature of roots, if roots are not given, do b2-4ac. when defining the features be careful about which b2-4ac to use
to find sum/product of roots (-b/a and c/a) the equation must be in standard form first
when the equation asks for a specific max/min value
solve for y value
when the problem asks for the time at which a max/min occurs
find x value
algabraically- use AOS, find x value, then plug into equation or use calc graph
“At what time does the max/min occur?”
if question asks for the time at which an object is at a gven height, look for x value. set equation equal to given height
if the questions asks for the time at which an object hits the ground, set equation to zero to find roots (or graph?)
(x-h)2 = 4p (y-k)
locus formula of a parabola
parabola= collection of all the points equidistant from focus & directrix
when solving, change from locus to standard form (then you can check with calc)
positive vs negative leading coefficient
if there is a negative leading coefficient, the locus equation NEEDS to be negative. so the equation becomes -(x-h)2 = 4p (y-k)
completing the square: any leading coefficients in x2 are factored out from the start, so the same value is added in the parenthesis and subtracted out of it. the constant is added to the side of 0 so the (x + 1)2= that number, then it’s squared, so the final answer is that x equals something. it also FOLLOWS THROUGH WITH THE b/2 squared thing.
steps to writing quadratic equations in vertex form by completing the square
if there’s a leading coefficient, it’s simply placed outside the parenthesis and the numbers inside the parenthesis are adjusted to it. the third value is taken from the 2nd one (b/2)2, and depending on how big the value is outside the parenthesis, the constant number (that ‘ll be subtracting the number) is adjusted (unlike completing the square where it stays constant) after, the equation is set up so the leading coefficient is outside the parenthesis, then (x, then plus or minus based on the first equation, then the squared value of the third number in the parenthesis, so for example the end result looks like 2(x + 2)2 + 4 and the vertex would be -2 and 4.
changing from standard to vertex form (NOT completing the square)
the final answer is expressed as y= (x-h)2 + k and vertex stated as (h,k)
use b2-4ac for roots (when graph touches x axis) and -b/2a for AOS, finding x value for a vertex
Nature of roots
Discriminant
Algabraically finding the set of x intercepts for parabolas
B2-4ac
