Question

What is the equation of the parabola that has a vertex at `(-4, 121)` and passes through point `(7,0)`?

Answer 1

`y=-(x+4)^2+121`

Explanation:

Given vertex at `(-4, 121)` and a point `(7, 0)`
`h=-4`
`k=121`
`x=7`
`y=0`

Use the standard form. Substitute the values to solve for `p`.

`(x-h)^2=-4p(y-k)`

`(7--4)^2=-4p(0-121)`

`(11)^2=-4p(-121)`
`121=4(121)p`

`121/121=(4(121)p)/121`

`cancel121/cancel121=(4(cancel121)p)/cancel121`

`1=4p`

`p=1/4`

the equation is now

`(x--4)^2=-4(1/4)(y-121)`

`(x+4)^2=-1(y-121)`

`(x+4)^2=-y+121`

`y=-(x+4)^2+121`

graph{y=-(x+4)^2+121[-100,300,-130,130]}

Have a nice day !! from the Philippines.