Question

How do you find the vertex and the intercepts for `y=-x^2+x+12`?

Answer 1

Vertex: `(1/2,49/4)`
y-intercept: `12`
x-intercepts: `4` and `(-3)`

Explanation:

The vertex form of a parabola is
`color(white)("XXX")y=color(m)(x-color(blue)(a))+color(red)(b)`
with vertex at `(color(blue)(a),color(red)(b))`

Converting the given `y=-x^2+x+12` into vertex form:

`color(white)("XXX")y=color(green)((-1))(x^2-x+(1/2)^2)+12 + (1/2)^2`

`color(white)("XXX")y=color(green)((-1))(x-color(blue)(1/2))^2+color(red)(49/4)`

giving us the vertex at `(color(blue)(1/2,)color(red)(49/4))`

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The y-intercept is the value of `y` when `color(magenta)(x=0)`

`y=-x^2+x+12` when `color(magenta)(x=0)`
`rarr y=-color(magenta)(0)^2+color(magenta)(0)+12=12`

So the y-intercept is `0`

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The x-intercepts are the values of `x` when `color(brown)(y=0)`

`y=-x^2+x+12` when `color(brown)(y=0)`
`rarr -x^2+x+12=color(brown)(0) color(white)("XX")orcolor(white)("XX")x^2-x-12=0`

This can be factored as
`color(white)("XXX")(x-4)(x+3)=0`
`rarr x=4color(white)("XX")orcolor(white)("XX")x=-3`

So the x-intercepts are `{4,-3}`

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Here is the graph of this relation for verification purposes:
graph{-x^2+x+12 [-10.29, 18.19, -1.36, 12.9]}