Question

Evaluate h(x)=12+x2
for x=−5

Answer 1

First, you need to evaluate the function correctly.

`\tt \: h(x) = 12 \: + {x}^(2)`

• when x = -5, we need to substitute -5 into the function

So ,

`\tt \: h(-5) = 12 + (-5)^2`

Now, solve the equation

`\tt \: {h(-5)= 12 + (-5)²} \\ \tt \: {h(-5) = 12 + (25)} \\ \tt \: {h(-5) = \green{37}}`

Therefore, when x is equal to -5, the value of h(x) is 37.