anonymous
  • anonymous
find the x-intercepts of the parabola with vertex (4,-1) and y-intercept (0,15) write your answer in this form:(x1,y1),(x2,y2) f necessary, round to the nearest Hundredth
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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campbell_st
  • campbell_st
well the vertex form of a parabola is \[y=a(x - h)^2 + k\] where (h, k) is the vertex. so substitute you vertex values, h = 4 and k = -1 into the equation. tThe y-intercept is on the parabola... so to find the value of a you need to substitute x = 0 and y = 15 and solve for a. then you can rewrite the equation of the parabola in standard form that's the 1st part hope it makes sense
campbell_st
  • campbell_st
once you have the equation in standard form you may need to use the general quadratic formula for the x intercepts
anonymous
  • anonymous
soo would it be y=x^2-8x+15

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campbell_st
  • campbell_st
that's the correct equation... and the equation can be factored.
anonymous
  • anonymous
its wrong
campbell_st
  • campbell_st
so what did you get for the x-intercepts
anonymous
  • anonymous
-4
campbell_st
  • campbell_st
the equation is \[y=x^2 - 8x + 15~~~~then~~~~y=(x -5)(x-3)\]
campbell_st
  • campbell_st
so the intercepts are
campbell_st
  • campbell_st
where y = 0 so solve x - 5 = 0 and x - 3 = 0
anonymous
  • anonymous
am confused
campbell_st
  • campbell_st
ok.... so the x-intercepts these are points on the x-axis so y = 0 if you have a parabola to find the x-intercepts let y = 0 and solve for x so \[0 = x^2 - 8x + 15~~~or~~~0 =(x -5)(x -3)\] if either of the factors (x -5) or (x -3) are equal to zero the equation is equal to zero... so then you need to let the factors equal zero and solve for x so solve x - 5 = 0 and x - 3 = 0 what would you have...?
campbell_st
  • campbell_st
the other thing to think about is that the x-intercepts are the same distance from the x value in the vertex .
anonymous
  • anonymous
am go fail this test
campbell_st
  • campbell_st
so if you the solve the 2 equations what do you get x - 5 = 0 and x - 3 = 0
campbell_st
  • campbell_st
these values are the x-intercepts
anonymous
  • anonymous
x=5,3
campbell_st
  • campbell_st
correct so the intercepts are at (3,0) and (5, 0) the vertex has an x value of x = 4... the intercepts are the same distance from this value
anonymous
  • anonymous
thank u soo much
campbell_st
  • campbell_st
yw
anonymous
  • anonymous
it was wrong
campbell_st
  • campbell_st
so did you write the intercepts as (3, 0) and (5, 0)..?
anonymous
  • anonymous
yea
campbell_st
  • campbell_st
here is a graph of the parabola you gave... and it meets all the conditions... can you check the vertex is (4, -1) and y-intercept is y = 15 or (0. 15)
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