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zimovet [89]
3 years ago
13

​ f(x)=x^2−8x+19 ​ . What is the minimum value of the function?​

Mathematics
2 answers:
topjm [15]3 years ago
8 0

Answer:

Step-by-step explanation

I took the test its        ANS: 3

vitfil [10]3 years ago
7 0
The minimum value for the function is 19 because if the x is 0 then the f(x) is 19. Therefore 19 is the lowest.
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now  \bf \boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}\qquad\qquad  vertex\ ({{ h}},{{ k}})\\\\
-----------------------------\\\\
y=a(x-0)^2+0\implies y=ax^2
\\\\\\
\textit{what's the value of "a"?  well, we know }
\begin{cases}
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\end{cases}
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