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If someone could help me work through this problem I would appreciate it! Consider the quadratic function

$f(x) = a + (bx − 1)^2$

where $a, b$ are arbitrary constants.

(a) Find the minimum value of f and the x for which it is achieved.

(b) For what values of a and b will the equation $f(x) = 0$ have two real solutions? Thank you.

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  • $\begingroup$ Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? And to help you: what is the minimum value of $x^2$ for real values of $x$? $\endgroup$ – Rory Daulton Mar 11 '15 at 22:25
  • $\begingroup$ a) note that $(bx-1)^2\geq 0$ for every $x$ so that $f(x)\geq a$ for every $x$. Can you find $x$ such that $(bx-1)^2=0$? $\endgroup$ – Surb Mar 11 '15 at 22:28
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a)If we take the derivative we find that the minimum/maximum is at points where the derivative is 0. $$0=2(bx-1)b=2b^2x-2b=2bx-2\implies bx=1\implies x=\frac 1b (b\neq 0)$$ Now we plug this into the equation: $$f_{min}= a+(1-1)^2=a$$ So the point is $\left(\frac 1b, a\right)$ Or if $b=0,$ it is $\left(any \, point,a+1\right)$

b) Expanding the quadratic gives $$a+(1-bx)^2=a+1-2bx+b^2x^2=b^2x^2-2bx+(a+1)$$ Now the discriminant has to be greater than 0 for it to have 2 real solutions: $$D=4b^2-4b^2(a+1)=4b^2(1-a+1)=4b^2a>0\implies b^2a>0$$ Dividing that by $a$ gives $b\neq 0$, and dividing it by $b^2$ gives $a>0$

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  • $\begingroup$ Your solution to part a) is correct only if $b\ne0$. Also, since the tag is (algebra-Precalculus) the OP probably wants a non-calculus solution. $\endgroup$ – Rory Daulton Mar 11 '15 at 22:42

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