Question

Find the x-intercept of the parabola with vertex (4,75) and y-intercept (0,27)
(x 1 ,y 1 ),(x 2 ,y 2 )

Answer 1

\(y\) intercept is what you get when you replace \(x\) by \(0\)
if your parabola is \(y=ax^2+bx+c\) the the \(y\) intercept is \(c\) which you know is \(27\)

Answer 2

since the vertex is \((4,75)\) you know it looks like
\[y=a(x-4)^2+75\] and now we can solve for \(a\) by replacing \(x\) by \(0\) and setting the result equal to \(27\)

Answer 3

you get
\[27=a(0-4)^2+75\]
\[27=16a+75\]
\[-48=16a\]
\[-4=a\]

Answer 4

damn that is wrong sorry, should be \(-3=a\)

Answer 5

therefore
\[y=-3(x-4)^2+75\] and now we can set it equal to zero and solve

Answer 6

you get
\[-3(x-4)^2+75=0\]
\[-3(x-3)^2=-75\]
\[(x-3)^2=25\]
\[x-3=\pm5\]
\[x=8\] or \[x=-2\]

Answer 7

and the \(x\) intercepts are therefore \((-2,0)\) and \((8,0)\)