anonymous
  • anonymous
Find the x intercepts of the parabola with vertex (1,-17) and y intercept (0,-16)
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
please help me I will fan and medal
anonymous
  • anonymous
http://www.mathwarehouse.com/geometry/pa... Vertex Form of Equation y= a(x-h)^2 + k Knowing (h,k) is the vertex, and the vertex is (1,-17) y = (x-1)^2 - 17 y = x^2 - 2x + 1 - 17 y = x^2 - 2x - 16 Knowing: y = ax 2 + bx + c a = 1 b = -2 c = -16 Solving for x intercepts: http://www.analyzemath.com/quadratics/ve... x1 = [ - b + sqrt (b^2 - 4 a c) ] / 2 a x2 = [ - b - sqrt (b^2 - 4 a c) ] / 2 a Solving: x1 = [2 + sqrt(4 + 64)] / 2 x1 = (2 + sqrt(68)) / 2 x2 = [ - b - sqrt (b^2 - 4 a c) ] / 2 a x2 = [2 - sqrt(4 + 64)] / 2 x2 = (2 - sqrt(68)) / 2 Therefore, rounding to the nearest hundredth: We know y is zero because they are x-intercepts. (5.12, 0), (-3.12, 0)
anonymous
  • anonymous
so all of that is rounded to the nearest hundredth?

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anonymous
  • anonymous
@SpooderWoman
anonymous
  • anonymous
yep
anonymous
  • anonymous
sorry it took so long to reply I had to take my dogs out :p
anonymous
  • anonymous
that's okay :) can you help me do the same thing with vertex -1,-16 and y intercept 0,-15?
anonymous
  • anonymous
Sure!
anonymous
  • anonymous
Okay: first, write the equation of the parabola in the required form: (y - k) = a·(x - h)² Here, (h, k) is given as (-1, -16). So you have: (y + 16) = a · (x + 1)² Unfortunately, a is not given. However, you do know one additional point on the parabola: (0, -15): -15 + 16 = a· (0 + 1)² .·. a = 1 .·. the equation of the parabola in vertex form is y + 16 = (x + 1)² The x-intercepts are the values of x that make y = 0. So, let y = 0: 0 + 16 = (x + 1)² 16 = (x + 1)² We are trying to solve for x, so take the square root of both sides - but be CAREFUL! ± 4 = x + 1 ...... remember both the positive and negative roots of 16...... Solving for x: x = -1 + 4, x = -1 - 4 x = 3, x = -5. Or, if you prefer, (3, 0), (-5, 0).
anonymous
  • anonymous
Are you taking FLVS?
anonymous
  • anonymous
FLVS???
anonymous
  • anonymous
Online class
anonymous
  • anonymous
These seem like the questions I had on my virtual school :p
anonymous
  • anonymous
Yes I'm taking summer courses with Ed options Academy trough Plato
anonymous
  • anonymous
Nice
anonymous
  • anonymous
Any more questions you need help with?
anonymous
  • anonymous
Hold on for a sec?
anonymous
  • anonymous
Yep :)
anonymous
  • anonymous
vertex 1,245 and y intercept 0,240
anonymous
  • anonymous
y = a(x-1)² + 245 240 = a+245 a = -5 y = -5(x-1)² + 245 = 0 -5(x² - 2x + 1) + 245 = 0 x² - 2x - 48 = 0 (x-8)(x+6) = 0 x = -6, 8 x-intercepts: (-6,0) and (8,0)
anonymous
  • anonymous
Want me to explain that one more?
anonymous
  • anonymous
no I think I got that one
anonymous
  • anonymous
Alright :) (Thanks for the testimony)
anonymous
  • anonymous
no problem :) vertex (1,1) y intercept (0,-3)
anonymous
  • anonymous
vertex form of equation for a parabola: y=A(x-h)^2+k, (h,k)=coordinates of vertex solving for A: -3=A(0-1)^2+1 A=-4 equation: y=-4(x-1)^2+1 x-intercepts set y=0 -4(x-1)^2+1=0 (x-1)^2=1/4 x-1=±√(1/4)=±1/2 x=1±1/2 x-intercepts: x=1/2 x=3/2
anonymous
  • anonymous
so how w\[(x _{1},y _{1}),(x _{2},y _{2})\]ould you write that as
anonymous
  • anonymous
For which one
anonymous
  • anonymous
for the one you just did
anonymous
  • anonymous
oh I think it would be (3/2,0) (0,-3) To show the x-intercept and the y-intercept.
anonymous
  • anonymous
okay thank you that was my last one
anonymous
  • anonymous
No problem! Good Luck with your schooling!

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