Question

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

Answer 1

Answer:

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

Step-by-step 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:

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

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

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:

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

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

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

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