Question

53. Given the function $ p(x)=\left\{\begin{array}{ll}x^{2}+5 x+6, & x<-4 \\ x+4, & x \geq-4\end{array}\right. $, what is the range when the domain is $ \{-6,-4,-2,0,2\} $ ? (4) $ \{-60,0,2,4,6\} $ (B) $ \{-2,0,2,4,6\} $ (c) $ \{0,2,6,12,20\} $ (2) $ \{0,2,4,6,12\} $

          53. Given the function \( p(x)=\left\{\begin{array}{ll}x^{2}+5 x+6, & x<-4 \\ x+4, & x \geq-4\end{array}\right. \), what is the range when the domain is \( \{-6,-4,-2,0,2\} \) ?
(4) \( \{-60,0,2,4,6\} \)
(B) \( \{-2,0,2,4,6\} \)
(c) \( \{0,2,6,12,20\} \)
(2) \( \{0,2,4,6,12\} \)

$53. Given the function p(x)={ x^2+5 x+6, x<-4 x+4, x ≥-4 ., what is the range when the domain is {-6,-4,-2,0,2} ? (4) {-60,0,2,4,6} (B) {-2,0,2,4,6} (c) {0,2,6,12,20} (2) {0,2,4,6,12}$

Added by Tammy J.

Precalculus

Jay Abramson 1st Edition

Chapter 3

Instant Answer

Solved by Expert Suman Saurav Thakur

Step 1

- For $x = -6$, use $x^2 + 5x + 6$ because $-6 < -4$. - For $x = -4$, use $x + 4$ because $-4 \geq -4$. - For $x = -2$, use $x + 4$ because $-2 \geq -4$. - For $x = 0$, use $x + 4$ because $0 \geq -4$. - For $x = 2$, use $x + 4$ Show more…

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53. Given the function $ p(x)=\left\{\begin{array}{ll}x^{2}+5 x+6, & x<-4 \\ x+4, & x \geq-4\end{array}\right. $, what is the range when the domain is $ \{-6,-4,-2,0,2\} $ ? (4) $ \{-60,0,2,4,6\} $ (B) $ \{-2,0,2,4,6\} $ (c) $ \{0,2,6,12,20\} $ (2) $ \{0,2,4,6,12\} $