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?

Mathematics

1 answer:

Leokris [45]3 years ago

6 0

Answer:

b = $\frac{1}{2}$

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 = $\frac{1}{2}$

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anzhelika [568]

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Step-by-step explanation:

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