Question

Write the equation of the parabola that has its x-intercepts at (1+
`√(5)`, 0) and (1-
`√(5)`,0) and passes through the point (4,8)

please put it in the form y = ...

thank you

Answer 1

`y=2x^2-4x-8`

Explanation:

Factored form of a parabola

`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+√(5)))(x-(1-√(5)))`

`\implies y=a(x-1-√(5))(x-1+√(5))`

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

`\implies a(4-1-√(5))(4-1+√(5))=8`

`\implies a(3-√(5))(3+√(5))=8`

`\implies4a=8`

`\implies a=2`

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

`\implies y=2(x-1-√(5))(x-1+√(5))`

Expand so that the equation is in standard form:

`\implies y=2(x^2-x+√(5)x-x+1-√(5)-√(5)x+√(5)-5)`

`\implies y=2(x^2-x-x+√(5)x-√(5)x+√(5)-√(5)+1-5)`

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

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