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

1 Answer
Oct 7, 2015

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