414,389 views
24 votes
24 votes
Y=X²-4x-5Domain-Range=Function=

User Harshal Deore
by
2.1k points

1 Answer

13 votes
13 votes

In this problem we have a quadratic equation

y=X²-4x-5

The domain of a quadratic equation is all real numbers

To find out the range we need to calculate the vertex of the parabola

Convert the quadratic equation into vertex form

so

(y-k)=(x-h)^2

y=X²-4x-5

y+5=x^2-4x

complete the square

y+5=(x^2-4x+4)-4

y+5+4=(x-2)^2

y+9=(x-2)^2

the vertex is the point (2,--9)

The quadratic equation represent a vertical parabola open upwards

so the range is the interval

{-9, infinite)

y-1=(x-2)^2 -------> is written as vertex form

The vertex of the parabola is the point (h,k)

(y-k)=(x-h)^2

using a graphing tool to better understand the problem

The range is the interval {-9, infinite)

The domain is the interva (-infinite, infinite)

the function is

y+9=(x-2)^2

f(x)=(x-2)^2-9

or

f(x)=x^2-4x-5

Y=X²-4x-5Domain-Range=Function=-example-1
User Rufflewind
by
2.6k points