Question

How do you graph the parabola `y=-x^2+x+12` using vertex, intercepts and additional points?

Answer 1

Follow the process below to find the key points:

Explanation:

`y=-x^2+x+12`

Let `x=0, y=12`
`:.` parabola passes through `(0, 12)`

Let `y=0`
`0=-x^2+x+12`
`0=x^2-x-12`
`0=(x-4)(x+3)`
`x=4` or `x=-3`
`:.` passes through `(4,0)` and `(-3, 0)`

This should be enough to sketch a graph, but since the question asks for the vertex (turning point), we should find this too. You can do this by completing the square, or differentiation.

`y=-x^2+x+12`
`y=-(x^2-x-12)`
`y=-([x-1/2]^2-1/4-12)`
`y=-([x-1/2]^2-49/4)`
`y=-(x-1/2)^2+49/4`
`:. "vertex"=(1/2, 49/4)`

`Or`

`y=-x^2+x+12`
`dy/dx=-2x+1`
Let `dy/dx=0`
`-2x+1=0`
`-2x=-1`
`x=1/2`
Let `x=1/2`
`y=-(1/2)^2+(1/2)+12`
`y=-1/4+1/2+12`
`y=49/4`
`:.` vertex`=(1/2, 49/4)`

You only need one of these methods, but you may want to use the other one to check you haven't made any mistakes (I managed to catch quite a few mistakes just now from this).

Now, we can draw the graph

graph{-x^2+x+12 [-10, 10, -5, 5]}