Question

How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of `y= -x^2-8x+10`?

Answer 1

Vertex: `(-4,26))`
Axis of symmetry: `x=-4`
y-intercept: `10`
x-intercepts: `-4+-sqrt(26)`
Domain: `RR`
Range: `RR nn (-oo,26]`

Explanation:

Vertex
Re-writing `y=-x^2-8x+10` in vertex form:
`color(white)("XXX")y= (-1)(x^2+8x+4^2)+10+16)`
`color(white)("XXX")y=(-1)(x-(color(red)(-4)))^2+(color(blue)(26))`
with vertex at `(color(red)(-4),color(blue)(26))`

Axis of Symmetry
Any parabola in the form `y=ax^2+bx+c` has as an axis of symmetry the vertical line passing through the vertex.

In this case `x=-4`

y-intercept
The y-intercept is the value of `y` when `x=0`
`color(white)("XXX")y=-(0)^2-8(0)+10 = 10`

x-intercepts
The x-intercepts are the values of `x` when `y=0`
`0=(-1)(x+4)^2+26`
`(x+4)^2=26`
`x+4=+-sqrt(26)`
`x=-4+-sqrt(26)`

Domain
`y=-x^2-8x+10` is defined for all Real values of `x`

Range
Since a parabolic equation with a negative coefficient for the `x^2` term opens downward, the vertex gives a maximum value for the Range.
The Range is `<=26`