<span>Minimum wykresu funkcji kwadratowej znajduje się w ( -1, 2). Punkt ( 2 , 20) jest również od paraboli. Która funkcja reprezentuje sytuację?
</span><span>Canonical form of the function
</span>f(x) = a* (x - p)² + q

A .f(x) = (x + 1)² + 2 ⇔ p= -1 , q = 2
B. f(x) = (x – 1)² + 2 we reject
C. f(x) = 2(x + 1)² + 2 ⇔ p = -1 , q = 2
D .f(x) = 2(x – 1)² + 2 we reject

The point (2,20) substitute
A f(x) = (x +1)² + 2
20 = (2 + 1 )² + 2
20 ≠ 9 +2
20 ≠ 11 we reject

D f(x) = 2* (x + 1)² + 2
20 = 2* (2+1)² + 2
20 = 2 * 3² + 2
20 = 2 * 9 + 2
20 = 18 + 2

Reply C

8 0

4 years ago

Trigonometric Identities and Applications?

alex41 [277]

21)

$\bf 8cos^2(\theta )-3cos(\theta )=0\implies \stackrel{common~factor}{cos(\theta )}[8cos(\theta )-3]=0\\\\
-------------------------------\\\\
cos(\theta )=0\implies \measuredangle \theta =cos^{-1}(0)\implies \measuredangle \theta =90^o\\\\
-------------------------------\\\\
8cos(\theta )-3=0\implies 8cos(\theta )=3\implies cos(\theta )=\cfrac{3}{8}
\\\\\\
\measuredangle \theta =cos^{-1}\left( \cfrac{3}{8}\right)$

22)

$\bf \begin{cases}
y=cos(2x)\\
y=cos^2(x)-1
\end{cases}\implies \stackrel{intersection~at}{cos(2x)=cos^2(x)-1}
\\\\\\
2cos^2(x)-1=cos^2(x)-1\implies 2cos^2(x)-1-cos^2(x)+1=0
\\\\\\
2cos^2(x)-cos^2(x)=0\implies cos^2(x)=0\implies cos(x)=0
\\\\\\
\measuredangle \theta =cos^{-1}(0)\implies \measuredangle \theta =180^o~~,~~270^o$

4 0

3 years ago