# Graph square root of 8x+5 8x+5
Find the domain for y=8x+5 so that a list of x values can be picked to find a list of points, which will help graphing the radical.
Set the radicand in 2x greater than or equal to 0 to find where the expression is defined.
2x≥0
Divide each term by 2 and simplify.
Divide each term in 2x≥0 by 2.
2×2≥02
Cancel the common factor of 2.
Cancel the common factor.
2×2≥02
Divide x by 1.
x≥02
x≥02
Divide 0 by 2.
x≥0
x≥0
The domain is all values of x that make the expression defined.
Interval Notation:
[0,∞)
Set-Builder Notation:
{x|x≥0}
Interval Notation:
[0,∞)
Set-Builder Notation:
{x|x≥0}
To find the radical expression end point, substitute the x value 0, which is the least value in the domain, into f(x)=22x+5.
Replace the variable x with 0 in the expression.
f(0)=22(0)+5
Simplify the result.
Simplify each term.
Multiply 2 by 0.
f(0)=20+5
Rewrite 0 as 02.
f(0)=202+5
Pull terms out from under the radical, assuming positive real numbers.
f(0)=2⋅0+5
Multiply 2 by 0.
f(0)=0+5
f(0)=0+5
Add 0 and 5.
f(0)=5
The final answer is 5.
5
5
5
The radical expression end point is (0,5).
(0,5)
Select a few x values from the domain. It would be more useful to select the values so that they are next to the x value of the radical expression end point.
Substitute the x value 1 into f(x)=22x+5. In this case, the point is (1,22+5).
Replace the variable x with 1 in the expression.
f(1)=22(1)+5
Simplify the result.
Multiply 2 by 1.
f(1)=22+5
The final answer is 22+5.
y=22+5
y=22+5
y=22+5
Substitute the x value 2 into f(x)=22x+5. In this case, the point is (2,9).
Replace the variable x with 2 in the expression.
f(2)=22(2)+5
Simplify the result.
Simplify each term.
Multiply 2 by 2.
f(2)=24+5
Rewrite 4 as 22.
f(2)=222+5
Pull terms out from under the radical, assuming positive real numbers.
f(2)=2⋅2+5
Multiply 2 by 2.
f(2)=4+5
f(2)=4+5
Add 4 and 5.
f(2)=9
The final answer is 9.
y=9
y=9
y=9
The square root can be graphed using the points around the vertex (0,5),(1,7.83),(2,9)
xy0517.8329
xy0517.8329
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