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).

Answer 1

so since we know its x-intercepts or namely zeros or roots, we can reword all that to

what's the equation of a quadratic with roots of 1+√5 and 1-√5 and passes through the point (4, 8)?

`\begin{cases} x = 1+√(5) &\implies x -1-√(5)=0\\ x = 1-√(5) &\implies x -1+√(5)=0 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{original~polynomial}{a ( x -1-√(5) )( x -1+√(5) ) = \stackrel{0}{y}} \\\\[-0.35em] ~\dotfill\\\\ ( x -1-√(5) )( x -1+√(5) )\implies\stackrel{ \textit{difference of squares} }{[(x-1)-√(5)][(x-1)+√(5)]}`

`(x-1)^2-(√(5))^2\implies (x-1)^2-5\implies (x^2-2x+1)-5\implies x^2-2x-4 \\\\[-0.35em] ~\dotfill\\\\ a ( x -1-√(5) )( x -1+√(5) ) =y\implies a(x^2-2x-4)=y \\\\\\ \textit{we also know that } \begin{cases} x=4\\ y=8 \end{cases}\implies a( ~~(4)^2-2(4)-4 ~~ )=8 \\\\\\ a(16-8-4)=8\implies 4a=8\implies a=\cfrac{8}{4}\implies \boxed{a=2} \\\\[-0.35em] ~\dotfill\\\\ a(x^2-2x-4)=y\implies 2(x^2-2x-4)=y\implies {\Large \begin{array}{llll} 2x^2-4x-8=y \end{array}}`

Check the picture below.