Question

find f(-1) & f(-5) for f(x)= x^2+6x-5​

Answer 1

$\sf \bf {\boxed {\mathbb {GIVEN:}}}$

♨$\:f(x) = {x}^{2} + 6x - 5$

$\sf \bf {\boxed {\mathbb {TO\:FIND:}}}$

☎$\:(i)\:f( - 1) \: \: and \: \: (ii) \: f( - 5)$

$\sf \bf {\boxed {\mathbb {SOLUTION:}}}$

For finding $f(-1)$ and $f(-5)$, we just have to plug in $'-1'$ and $'-5'$ for all values of $x$.

And so we have,

$i) \:  f( - 1) = ( { - 1})^{2} + 6( - 1) - 5$

$= ( - 1 \times - 1) - 6 - 5$

$= 1 - 6 - 5$

$= 1 - 11$

$= - 10$

$ii) \: f( - 5) = ({ - 5})^{2} + 6( - 5) - 5$

$= ( - 5 \times - 5) - 30 - 5$

$= 25 - 30 - 5$

$= 25 - 35$

$= - 10$

Therefore, the values of $f(-1)$ and $f(-5)$ are $\boxed{ -10 }$ and $\boxed{ -10 }$ respectively.

$\sf\red{Hope\:it\:helps.}$

Answer 2

Answer:

\:f(x) = {x}^{2} + 6x - 5f(x)=x

2

+6x−5

\sf \bf {\boxed {\mathbb {TO\:FIND:}}}

TOFIND:

\:(i)\:f( - 1) \: \: and \: \: (ii) \: f( - 5)(i)f(−1)and(ii)f(−5)

\sf \bf {\boxed {\mathbb {SOLUTION:}}}

SOLUTION:

For finding f(-1)f(−1) and f(-5)f(−5) , we just have to plug in '-1'

′

−1

′

and '-5'

′

−5

′

for all values of xx .

And so we have,

i) \: f( - 1) = ( { - 1})^{2} + 6( - 1) - 5i) f(−1)=(−1)

2

+6(−1)−5

= ( - 1 \times - 1) - 6 - 5=(−1×−1)−6−5

= 1 - 6 - 5=1−6−5

= 1 - 11=1−11

= - 10=−10

ii) \: f( - 5) = ({ - 5})^{2} + 6( - 5) - 5ii)f(−5)=(−5)

2

+6(−5)−5

= ( - 5 \times - 5) - 30 - 5=(−5×−5)−30−5

= 25 - 30 - 5=25−30−5

= 25 - 35=25−35

= - 10=−10

Therefore, the values of f(-1)f(−1) and f(-5)f(−5) are \boxed{ -10 }

−10

and \boxed{ -10 }

−10

respectively.