Question

How do you find the vertex and intercepts for `y=-x^2-2x+3`?

Answer 1

Vertex (-1, 4)
x-intercepts: x = 1 and x = -3

Explanation:

x-coordinate of vertex:
`x = -b/(2a) = 2/-2 = - 1`
y-coordinate of vertex:
y(-1) = - 1 + 2 + 3 = 4
Vertex (-1, 4).
To find y-intercept, make x = 0 --> y = 3
To find x-intercepts, make y = 0, and solve:
`- x^2 - 2x + 3 = 0`
Since a + b + c = 0, use shortcut.
The 2 real roots (x-intercepts) are:
x = 1 and `x = c/a = 3/(-1) = - 3`
graph{- x^2 - 2x + 3 [-10, 10, -5, 5]}

Answer 2

The vertex is `(-1,4)`, and is the maximum point of the parabola.

The x-intercepts are `(1,0)` and `(-3,0)`.

The y-intercept is `(0,3)`.

Explanation:

`y=-x^2-2x+3` is the equation of a parabola in standard form: `y=ax^2+bx+c`, where `a=-1`, `b=-2`, and `c=3`.

The vertex of a parabola is the minimum or maximum point, `(x,y)`, of the parabola.

To determine the vertex of a parabola from the standard equation, use the following formulas:

`x=(-b)/(2a)`

`y=f(h)`.

Vertex of Parabola

`x=(-(-2))/(2*-1)`

`x=2/(-2)`

`x=-1`

To determine `y`, substitute the value for `x` into the standard equation and solve.

`y=f(h)=-(-1)^2-2(-1)+3`

`y=f(h)=-1+2+3`

`y=f(h)=4`

The vertex is `(-1,4)`, and is the maximum point of the parabola.

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`"x-intercepts"`

The x-intercepts are the values for `x` when `y=0`. Substitute `0` for `y`.

`-x^2-2x+3=0`

Find two numbers that when added equal `-2` and when multiplied equal `3`. `1` and `-3` meet the criteria.

`-(x-1)(x+3)=0`

Multiply both sides by `-1`.

`(x-1)(x+3)=0`

Solutions for `x`.

`(x-1)=0`

`x=1`

`(x+3)=0`

`x=-3`

The x-intercepts are `(1,0)` and `(-3,0)`.

Determine the y-intercept by substituting `0` for `x` in the standard equation.

`y=-1(0)-2(0)+3`

The y-intercept is `(0,3)`.

graph{y=-x^2-2x+3 [-10, 10, -5, 5]}