anonymous
  • anonymous
I need help with parabolas and solving quadratic-linear systems. 8th grade algebra one.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
whats the q?
amistre64
  • amistre64
from the sound of the statement you made, it seems like you need help finding where quads and lines intersect each other...
anonymous
  • anonymous
Yes but I don't understand the step in the process that is xorigin = -)b/2a) = 0. Does that have any significance and does it always equal 0?

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amistre64
  • amistre64
do you know the generic form for a quadratic equation? ax^2 +bx + c ?
amistre64
  • amistre64
And do you know the generic shape for a quadratic equation that is graphed? it looks like a "U" right?
anonymous
  • anonymous
I didn't know the generic form for a quadratic equation but I did know about the parabola. But these types of problems end up to be a parabola with a line intersecting it.
amistre64
  • amistre64
the formula (-b/2a) is gotten from the generic form of the quadratic equation; where a = some number; b = some number; and c = some number... for example: 2x^2 +3x +7 is a quadratic equation with a=2; b=3; and c = 7 the form -b/2a is the formula to find the "axis of symmetry" of the parabole; and it is the value of "x" that that "U" shape strats to bend around itself and go back up again; in this equation tha twould be: -3/2(2) ; which equals -3/4
amistre64
  • amistre64
if the parabold is positioned at the origin; then the form (-b/2a) will always be zero; but the parabola is not always positioned at the otigin :)
amistre64
  • amistre64
you gotta read thru the typos :)
anonymous
  • anonymous
Oh I get it! Thank you so much! Alright, so let's say I have two equations; y = x^2 and y = x +2 What would a and b and c be equal to?
anonymous
  • anonymous
thanks amistre for explaining it to her. sorry about the paula got sidetracked.
anonymous
  • anonymous
@abigail - it's fine
amistre64
  • amistre64
for the quadratic part; a = 1; b = 0 and c = 0; the linear part has different "constants" to it tho. what I think you are looking to find is the points that x^2 and x+2 have in common like this:
1 Attachment
amistre64
  • amistre64
abigail: :) is alright, you can still help if youd like ;)
anonymous
  • anonymous
Ya that's what I need to find ;) so when I'm solving this should I write the formula would -b/2a be equal to 0?
amistre64
  • amistre64
when we wnat to know the points that grpah have in common; we make their equations equal like this: x^2 = x+2 ; next we put everything to one side x^2 -x -2 = 0 ; next we solve for the values of "x" that make this equation equal 0, and we do that easily by the quadratic formula. Or by factoring the equation. the quadratic formula is: -b/2a +- sqrt(b^2 -4(a)(c))/2a
amistre64
  • amistre64
x^2 -x -2; a=1, b=-1, c=-2 --1/2(1) +- sqrt((-1)^2 - 4(1)(-2))/2(1) 1/2 +- sqrt(1+8)/2 1/2 +- sqrt(9)/2 1/2 +- 3/2 x = 1/2 + 3/2 = 4/2 = 2 AND x = 1/2 - 3/2 = -2/2 = -1 the points they have in common are going to be: (-1,y1) (2,y2)
amistre64
  • amistre64
we find y1 and y2 by using the values for x that we found into one of the equations: x^2 is a good choice if you know your squares :) (-1)^2 = 1 (-1,1) is a point they share in common (2)^2 = 4 (2,4) is the other point they share in common
anonymous
  • anonymous
Gotchya :D Thanks sooo much!
anonymous
  • anonymous
id say you covered everything amistre i have nothing to input on this lol
amistre64
  • amistre64
:) mighta covered it all, but explaining it good enough is another matter :)
anonymous
  • anonymous
thats soo the truth either way well done!

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