Let b be a real number. For what value of b does the function f(x)=x^2-2bx+(b^2+2b-1) have one and only one x-intercept?

b =  

Step-by-step explanation:

Using the discriminant for nature of roots ( x- intercepts )

If b² - 4ac = 0 then roots are real and equal ( that is only 1 x- intercept )

Given

f(x) = x² - 2bx + b² + 2b - 1

with a = 1, b = - 2b and c = b² + 2b - 1, then

(- 2b)² - 4(b² + 2b - 1) = 0 ← distribute and simplify left side

4b² - 4b² - 8b + 4 = 0

- 8b + 4 = 0 ( subtract 4 from both sides )

- 8b = - 4 ( divide both sides by - 8 )

b =

step-by-step explanation:

sry but any answer choices

you ignore the remainder, therefore the answer is y = x+2

step-by-step explanation:

you have to divide the denominator by the numerator. you can use synthetic division or long division.

[-5] .     1 .     7 .     11

x + 2 + 1/x+5