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nikdorinn [45]
1 year ago
13

Write the equation of the parabola that has its x-intercepts at (1+

ula1" title="\sqrt{5}" alt="\sqrt{5}" align="absmiddle" class="latex-formula">, 0) and (1-\sqrt{5},0) and passes through the point (4,8)
please put it in the form y = ...

thank you
Mathematics
1 answer:
VladimirAG [237]1 year ago
6 0

Answer:

y=2x^2-4x-8

Step-by-step explanation:

<u>Factored form of a parabola</u>

y=a(x-p)(x-q)

where:

  • p and q are the x-intercepts.
  • a is some constant.

Given x-intercepts:

  • (1+√5, 0)
  • (1-√5, 0)

Therefore:

\implies y=a(x-(1+\sqrt{5}))(x-(1-\sqrt{5}))

\implies y=a(x-1-\sqrt{5})(x-1+\sqrt{5})

To find a, substitute the given point (4, 8) into the equation and solve for a:

\implies a(4-1-\sqrt{5})(4-1+\sqrt{5})=8

\implies a(3-\sqrt{5})(3+\sqrt{5})=8

\implies4a=8

\implies a=2

Therefore, the equation of the parabola in factored form is:

\implies y=2(x-1-\sqrt{5})(x-1+\sqrt{5})

Expand so that the equation is in standard form:

\implies y=2(x^2-x+\sqrt{5}x-x+1-\sqrt{5}-\sqrt{5}x+\sqrt{5}-5)

\implies y=2(x^2-x-x+\sqrt{5}x-\sqrt{5}x+\sqrt{5}-\sqrt{5}+1-5)

\implies y=2(x^2-2x-4)

\implies y=2x^2-4x-8

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